Exponential and Logarithmic Functions

Exponential Growth and Decay Mechanics describes situations where a quantity changes by the same percentage over equal time intervals. This is different from linear change, where the same amount is added or subtracted each time. In exponential situations, the amount of change depends on the current size of the quantity.

For example, if a population grows by 5% each year, the increase in the second year is larger than the increase in the first year, because 5% is taken from a larger amount. Likewise, if a car’s value decreases by 12% each year, the loss each year becomes smaller in dollars, even though the percent loss stays the same.

In this lesson, you will learn how to build exponential models, how to tell the difference between growth and decay, and how to distinguish between discrete compounding and continuous compounding.

1. The basic idea of exponential change

If a starting amount is multiplied by the same factor repeatedly, the model is exponential. The general form is

$$A(t)=A_0(b)^t$$

where:

• \(A_0\) is the initial amount,
• \(b\) is the growth or decay factor,
• \(t\) is time or the number of periods.

If \(b>1\), the function represents exponential growth. If \(0<b<1\), it represents exponential decay.

When a quantity changes by a rate \(r\), the factor is connected to the rate by:

• Growth: \(b=1+r\)
• Decay: \(b=1-r\)

Here, \(r\) must be written as a decimal. For example:

• 8% growth means \(r=0.08\), so factor \(=1.08\)
• 15% decay means \(r=0.15\), so factor \(=0.85\)

2. Constructing exponential models

To build a model, identify two things first:

1. the initial value, and
2. the percentage rate of change.

Then decide whether the rate is applied at separate time intervals or continuously.

For a quantity changing by a fixed percentage each period, the model is

$$A(t)=A_0(1+r)^t \quad \text{for growth}$$ $$A(t)=A_0(1-r)^t \quad \text{for decay}$$

This form is commonly used when time is counted in years, months, days, or other repeated intervals.

3. Discrete compounding

Discrete compounding means the change happens at specific intervals, such as yearly, quarterly, monthly, or daily. The amount is updated after each interval.

If a quantity starts at \(P\), grows at annual rate \(r\), and is compounded \(n\) times per year for \(t\) years, then

$$A(t)=P\left(1+\frac{r}{n}\right)^{nt}$$

This formula is used often in finance, but it also applies to any repeated percentage change happening in equal steps.

In this formula:

• \(P\) = initial amount
• \(r\) = annual rate as a decimal
• \(n\) = number of compounding periods per year
• \(t\) = time in years

Here are common values of \(n\):

• \(n=1\): annually
• \(n=2\): semiannually
• \(n=4\): quarterly
• \(n=12\): monthly
• \(n=365\): daily

4. Continuous compounding

Continuous compounding means the growth happens all the time rather than at separate intervals. This uses the natural base \(e\), where \(e\approx 2.718\).

The model for continuous growth or decay is

$$A(t)=A_0e^{rt}$$

where:

• \(A_0\) is the initial amount,
• \(r\) is the continuous growth or decay rate,
• \(t\) is time.

If \(r>0\), the quantity grows. If \(r<0\), the quantity decays.

This form is especially useful in science, population modeling, cooling, radioactive decay, and situations where change is happening continuously.

5. Discrete vs. continuous compounding

Both discrete and continuous models describe repeated percentage change, but they do so in slightly different ways.

• Discrete compounding: change happens at set times.
• Continuous compounding: change happens constantly.

Compare the formulas:

$$A(t)=P\left(1+\frac{r}{n}\right)^{nt} \quad \text{discrete}$$ $$A(t)=Pe^{rt} \quad \text{continuous}$$

As the number of compounding periods \(n\) becomes very large, the discrete model approaches the continuous model. That is why \(e\) appears naturally in continuous growth and decay.

6. Recognizing growth and decay from the formula

You can often tell whether a model shows growth or decay just by looking at the multiplier or exponent.

• In \(A(t)=A_0(1.12)^t\), the factor \(1.12>1\), so this is growth.
• In \(A(t)=A_0(0.93)^t\), the factor is between 0 and 1, so this is decay.
• In \(A(t)=A_0e^{0.04t}\), the exponent coefficient is positive, so this is growth.
• In \(A(t)=A_0e^{-0.18t}\), the exponent coefficient is negative, so this is decay.

7. Important modeling steps

When writing an exponential model from a word description, follow this process:

1. Find the initial amount.
2. Convert the percent rate to a decimal.
3. Decide whether the situation is growth or decay.
4. Decide whether the change is discrete or continuous.
5. Choose the correct formula and substitute the values.

A very common mistake is forgetting to convert a percent to a decimal. For example, 6% must be written as \(0.06\), not \(6\).

Another common mistake is using the wrong time unit. If the rate is annual, then time should usually be measured in years unless the formula has been adjusted properly.

Worked Example 1: Simple annual growth

A town has a population of 18,000 and grows by 3% each year. Write a model for the population after \(t\) years.

Step 1: Identify the initial amount.

\(A_0=18000\)

Step 2: Convert the rate to decimal form.

\(r=0.03\)

Step 3: Since this is growth, use \(1+r\).

Growth factor \(=1.03\)

Step 4: Write the model.

$$P(t)=18000(1.03)^t$$

This means the population is multiplied by \(1.03\) each year.

Worked Example 2: Exponential decay

A medicine starts at 250 mg in the bloodstream and decreases by 20% each hour. Write a model and find the amount after 4 hours.

Step 1: Initial amount:

\(A_0=250\)

Step 2: Rate:

\(r=0.20\)

Step 3: Since this is decay, use \(1-r\).

Decay factor \(=0.80\)

Step 4: Write the model.

$$A(t)=250(0.80)^t$$

Step 5: Find the amount after 4 hours.

$$A(4)=250(0.80)^4$$ $$A(4)=250(0.4096)=102.4$$

After 4 hours, about 102.4 mg remains.

Worked Example 3: Discrete compounding with more than one period per year

An investment of $4,000 earns 6% annual interest compounded monthly. Find the amount after 3 years.

Step 1: Use the discrete compounding formula:

$$A(t)=P\left(1+\frac{r}{n}\right)^{nt}$$

Step 2: Identify the values.

• \(P=4000\)
• \(r=0.06\)
• \(n=12\)
• \(t=3\)

Step 3: Substitute.

$$A(3)=4000\left(1+\frac{0.06}{12}\right)^{12\cdot 3}$$ $$A(3)=4000(1.005)^{36}$$

Step 4: Evaluate.

$$A(3)\approx 4000(1.19668)=4786.72$$

After 3 years, the investment is worth about $4,786.72.

Worked Example 4: Continuous compounding

An account contains $5,000 and grows continuously at a rate of 4.5% per year. Find the amount after 6 years.

Step 1: Use the continuous model:

$$A(t)=Pe^{rt}$$

Step 2: Identify the values.

• \(P=5000\)
• \(r=0.045\)
• \(t=6\)

Step 3: Substitute.

$$A(6)=5000e^{0.045\cdot 6}$$ $$A(6)=5000e^{0.27}$$

Step 4: Evaluate.

$$e^{0.27}\approx 1.30996$$ $$A(6)\approx 5000(1.30996)=6549.80$$

After 6 years, the account contains about $6,549.80.

8. Comparing discrete and continuous growth

Suppose $5,000 grows at 4.5% for 6 years.

• Monthly compounding gives

$$A=5000\left(1+\frac{0.045}{12}\right)^{72}\approx 6543.34$$

• Continuous compounding gives

$$A=5000e^{0.27}\approx 6549.80$$

The continuously compounded amount is slightly larger, because interest is being added at every instant rather than only once per month.

9. Interpreting the parameters in context

In an exponential model, each number has a meaning.

In \(A(t)=1200(1.08)^t\):

• 1200 is the starting amount,
• 1.08 means an 8% increase each time period,
• \(t\) is the number of time periods.

In \(A(t)=900e^{-0.12t}\):

• 900 is the initial amount,
• \(-0.12\) means continuous decay at a rate of 12% per unit of time,
• \(t\) is measured in the time units used in the situation.

Understanding the meaning of each part helps you choose the correct model and explain your answer clearly.

10. Common errors to avoid

• Using the percent instead of the decimal: 7% should be \(0.07\).
• Mixing up growth and decay: growth uses \(1+r\), decay uses \(1-r\).
• Ignoring compounding frequency: monthly compounding needs \(n=12\).
• Using the wrong formula: discrete and continuous models are not the same.
• Wrong time units: if the rate is yearly, make sure time is in years unless converted.

11. Quick check for understanding

Ask yourself these questions when you see a problem:

• What is the starting amount?
• Is the quantity increasing or decreasing?
• What is the percentage rate as a decimal?
• Does the change happen in separate intervals or continuously?
• Which formula matches the situation?

Brief Summary

Exponential growth and decay occur when a quantity changes by a constant percentage over equal intervals of time. For discrete change, use formulas like \(A(t)=A_0(1+r)^t\) or $$A(t)=P\left(1+\frac{r}{n}\right)^{nt}$$. For continuous change, use \(A(t)=A_0e^{rt}\). The key is to identify the initial value, the rate, whether it is growth or decay, and whether the compounding is discrete or continuous.

Introduction

In exponential functions, you may already know bases such as 2, 3, or 10. But in many real-life situations, especially those involving continuous growth or decay, one special number appears again and again: e.

The number e is called the natural base. Its approximate value is

$$e \approx 2.71828$$

This lesson will explain what e means, where it comes from, and how to use it in exponential models.

1. Where does the number e come from?

A classic way to understand e is through compound interest.

Suppose you invest $1 at 100% annual interest.

• If the interest is added once at the end of the year, you get:

$$1(1+1)=2$$

• If it is compounded twice a year, you get:

$$\left(1+\frac{1}{2}\right)^2=2.25$$

• If it is compounded 4 times a year, you get:

$$\left(1+\frac{1}{4}\right)^4 \approx 2.4414$$

• If it is compounded 12 times a year, you get:

$$\left(1+\frac{1}{12}\right)^{12} \approx 2.613$$

As the compounding happens more and more often, the amount approaches a limiting value:

$$\left(1+\frac{1}{n}\right)^n \to e \quad \text{as } n\to\infty$$

This means that e can be defined as

$$e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$

So, e is the number you get when growth is compounded continuously.

2. Why is e called the natural base?

The base e is called natural because it appears naturally in many processes:

• population growth
• radioactive decay
• cooling of objects
• continuous interest
• spread of substances or diseases

Whenever the rate of change depends on the current amount, functions with base e often model the situation well.

3. Exponential models with base e

A general exponential model using the natural base is

$$A(t)=A_0e^{kt}$$

where:

• \(A(t)\) is the amount after time \(t\)
• \(A_0\) is the initial amount
• \(k\) is the growth or decay constant
• \(t\) is time

If \(k>0\), the function models growth.

If \(k<0\), the function models decay.

4. Continuous compounding formula

For money growing with continuous compounding, the formula is

$$A=Pe^{rt}$$

where:

• \(P\) is the principal (starting amount)
• \(r\) is the annual interest rate written as a decimal
• \(t\) is time in years
• \(A\) is the final amount

This formula is different from ordinary compound interest:

$$A=P\left(1+\frac{r}{n}\right)^{nt}$$

As \(n\) becomes very large, this ordinary compound interest formula approaches

$$A=Pe^{rt}$$

5. Important facts about e

• \(e\approx 2.71828\)
• \(e^0=1\)
• \(e^1=e\)
• \(e^x\) represents natural exponential growth or decay
• The inverse of \(e^x\) is the natural logarithm, written \(\ln x\)

That means:

$$y=e^x \quad \Longleftrightarrow \quad x=\ln y$$

This relationship helps us solve equations involving e.

6. Worked Example 1: Approximating e from compound interest

Use the expression \(\left(1+\frac{1}{n}\right)^n\) to estimate \(e\) when \(n=10\) and \(n=100\).

Step 1: Substitute \(n=10\)

$$\left(1+\frac{1}{10}\right)^{10}=(1.1)^{10} \approx 2.5937$$

Step 2: Substitute \(n=100\)

$$\left(1+\frac{1}{100}\right)^{100}=(1.01)^{100} \approx 2.7048$$

Conclusion: As \(n\) gets larger, the value gets closer to

$$e\approx 2.71828$$

This shows how e comes from repeated compounding becoming continuous.

7. Worked Example 2: Continuous compounding

$2000 is invested at 5% annual interest, compounded continuously. Find the amount after 3 years.

Step 1: Write the formula

$$A=Pe^{rt}$$

Step 2: Substitute the values

Here, \(P=2000\), \(r=0.05\), and \(t=3\).

$$A=2000e^{0.05(3)}=2000e^{0.15}$$

Step 3: Calculate

$$A\approx 2000(1.1618)=2323.60$$

Answer: After 3 years, the investment is worth approximately $2323.60.

8. Worked Example 3: Natural growth model

A bacteria culture starts with 500 bacteria and grows according to

$$N(t)=500e^{0.4t}$$

Find the number of bacteria after 4 hours.

Step 1: Substitute \(t=4\)

$$N(4)=500e^{0.4(4)}=500e^{1.6}$$

Step 2: Calculate

$$e^{1.6}\approx 4.9530$$

$$N(4)\approx 500(4.9530)=2476.5$$

Answer: After 4 hours, there are about 2477 bacteria.

9. Worked Example 4: Solving for time using e and ln

A substance decays according to

$$A=800e^{-0.2t}$$

How long will it take for the amount to decrease to 300?

Step 1: Substitute \(A=300\)

$$300=800e^{-0.2t}$$

Step 2: Isolate the exponential part

$$\frac{300}{800}=e^{-0.2t}$$

$$0.375=e^{-0.2t}$$

Step 3: Take the natural logarithm of both sides

$$\ln(0.375)=\ln\left(e^{-0.2t}\right)$$

$$\ln(0.375)=-0.2t$$

Step 4: Solve for \(t\)

$$t=\frac{\ln(0.375)}{-0.2}$$

$$t\approx \frac{-0.9808}{-0.2}=4.904$$

Answer: It will take about 4.9 time units.

10. How to recognize when to use e

You should think about using e when a problem mentions:

• continuous compounding
• continuous growth or continuous decay
• a model of the form \(A=A_0e^{kt}\)
• natural logarithms \(\ln\)

In these situations, the natural base e is usually the best base to use.

11. Common mistakes to avoid

• Forgetting to change percent to decimal: 5% must be written as \(0.05\).
• Using the wrong formula: continuous compounding uses \(A=Pe^{rt}\), not \(A=P(1+r)^t\).
• Dropping parentheses: write \(e^{rt}\), not \(e^r t\).
• Rounding too early: keep calculator values until the final step.
• For decay, forgetting the negative sign: decay models use \(k<0\).

12. Key ideas to remember

• The number e is approximately \(2.71828\).
• It comes from the limit

$$e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$

• It models situations involving continuous change.
• The main exponential model is

$$A(t)=A_0e^{kt}$$

• For continuous interest, use

$$A=Pe^{rt}$$

• To solve equations involving \(e\), use the natural logarithm \(\ln\).

Brief Summary

The natural base e is a special irrational number that appears when growth happens continuously. It can be defined using the limit

$$e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$

and it is used in models such as

$$A(t)=A_0e^{kt}$$

for growth and decay, and

$$A=Pe^{rt}$$

for continuously compounded interest. Understanding e helps you model real-world situations and solve equations involving natural logarithms.

Graphs of Exponential Functions are important because they model many real-life situations such as population growth, radioactive decay, compound interest, and cooling. In this lesson, you will learn how to recognize and sketch exponential graphs, identify their key features, and understand how transformations change their shape and position.

An exponential function has the variable in the exponent. A basic exponential function can be written as

$$y=a\,b^x$$

where:

• \(a\) is the initial value or vertical stretch factor,
• \(b\) is the base, with \(b>0\) and \(b\neq 1\),
• \(x\) is the input variable.

The shape of the graph depends mainly on the base \(b\).

• If \(b>1\), the function shows exponential growth.
• If \(0<b<1\), the function shows exponential decay.

For example, \(y=2^x\) is a growth function, while \(y=\left(\frac{1}{2}\right)^x\) is a decay function.

1. The Basic Graphs

Let us begin with the parent function \(y=b^x\).

Every basic exponential graph has some important features:

• Its domain is all real numbers: \(( -\infty, \infty )\).
• Its range is \((0, \infty)\).
• It crosses the y-axis at \((0,1)\), because \(b^0=1\).
• It has a horizontal asymptote at \(y=0\).
• It does not cross the x-axis.

A horizontal asymptote is a horizontal line that the graph gets very close to but does not reach.

For \(y=2^x\):

• As \(x\) increases, the values of \(y\) increase quickly.
• As \(x\) becomes very negative, the graph gets closer and closer to \(y=0\).

For \(y=\left(\frac{1}{2}\right)^x\):

• As \(x\) increases, the values of \(y\) decrease toward 0.
• As \(x\) becomes very negative, the graph rises.

2. Plotting an Exponential Function

To sketch an exponential graph, it helps to make a small table of values. Choose easy x-values such as \(-2, -1, 0, 1, 2\), then calculate the corresponding y-values.

For example, for \(y=2^x\):

$$\begin{aligned} x=-2 &\Rightarrow y=2^{-2}=\frac14 \\ x=-1 &\Rightarrow y=2^{-1}=\frac12 \\ x=0 &\Rightarrow y=2^0=1 \\ x=1 &\Rightarrow y=2^1=2 \\ x=2 &\Rightarrow y=2^2=4 \end{aligned}$$

These points are:

• \((-2, \tfrac14)\)
• \((-1, \tfrac12)\)
• \((0,1)\)
• \((1,2)\)
• \((2,4)\)

Plotting these points gives a smooth curve that rises from left to right and approaches \(y=0\) on the left.

3. Y-Intercept and Horizontal Asymptote

When graphing exponential functions, two of the most important features are the y-intercept and the horizontal asymptote.

For a function in the form

$$y=a\,b^x+k$$

the horizontal asymptote is

$$y=k$$

This is because as the exponential part gets very small, the graph approaches the value \(k\).

To find the y-intercept, let \(x=0\):

$$y=a\,b^0+k=a+k$$

So the y-intercept is

$$\bigl(0, a+k\bigr)$$

4. Transformations of Exponential Graphs

Transformations help us understand how a graph changes from the parent function. A more general exponential function can be written as

$$y=a\,b^{(x-h)}+k$$

Each part changes the graph in a specific way.

• \(a\): vertical stretch or compression, and reflection in the x-axis if \(a<0\)
• \(h\): horizontal shift
• \(k\): vertical shift

Vertical shift: Adding \(k\) moves the graph up or down.

• \(y=2^x+3\) shifts the graph of \(y=2^x\) up 3 units.
• The horizontal asymptote changes from \(y=0\) to \(y=3\).

Horizontal shift: Replacing \(x\) with \(x-h\) moves the graph horizontally.

• \(y=2^{x-1}\) shifts the graph right 1 unit.
• \(y=2^{x+2}\) shifts the graph left 2 units.

Vertical stretch or compression: Multiplying by \(a\) changes the steepness.

• If \(|a|>1\), the graph is stretched vertically.
• If \(0<|a|<1\), the graph is compressed vertically.

Reflection in the x-axis: If \(a<0\), the graph is flipped upside down.

For example, \(y=-2^x\) reflects the graph of \(y=2^x\) across the x-axis. Its values are all negative, and it approaches \(y=0\) from below.

5. Growth and Decay from the Graph

You can tell whether an exponential function represents growth or decay by looking at the graph.

• If the graph rises from left to right, it is usually growth.
• If the graph falls from left to right, it is usually decay.

This matches the base:

• Growth: \(b>1\)
• Decay: \(0<b<1\)

Be careful when there is a reflection. For example, \(y=-3^x\) decreases from left to right, but this is caused by reflecting the growth graph across the x-axis.

6. Worked Examples

Example 1: Sketch the graph of \(y=3^x\)

Step 1: Make a table of values.

$$\begin{aligned} x=-2 &\Rightarrow y=3^{-2}=\frac19 \\ x=-1 &\Rightarrow y=3^{-1}=\frac13 \\ x=0 &\Rightarrow y=3^0=1 \\ x=1 &\Rightarrow y=3^1=3 \\ x=2 &\Rightarrow y=3^2=9 \end{aligned}$$

Step 2: Plot the points. The graph passes through \((0,1)\), rises quickly to the right, and approaches \(y=0\) to the left.

Step 3: State the features.

• Domain: all real numbers
• Range: \((0,\infty)\)
• y-intercept: \((0,1)\)
• Horizontal asymptote: \(y=0\)

This is an exponential growth graph because \(3>1\).

Example 2: Sketch the graph of \(y=\left(\frac{1}{4}\right)^x\)

Step 1: Make a table.

$$\begin{aligned} x=-2 &\Rightarrow y=\left(\frac14\right)^{-2}=16 \\ x=-1 &\Rightarrow y=\left(\frac14\right)^{-1}=4 \\ x=0 &\Rightarrow y=1 \\ x=1 &\Rightarrow y=\frac14 \\ x=2 &\Rightarrow y=\frac1{16} \end{aligned}$$

Step 2: Describe the graph. The graph passes through \((0,1)\), decreases from left to right, and approaches \(y=0\) as \(x\) increases.

Step 3: State the features.

• Domain: all real numbers
• Range: \((0,\infty)\)
• y-intercept: \((0,1)\)
• Horizontal asymptote: \(y=0\)

This is exponential decay because \(0<\frac14<1\).

Example 3: Find the key features of \(y=2^{x}+5\)

This graph is the parent function \(y=2^x\) shifted up 5 units.

Horizontal asymptote: Since the graph moved up 5, the asymptote becomes

$$y=5$$

y-intercept: Let \(x=0\).

$$y=2^0+5=1+5=6$$

So the y-intercept is \((0,6)\).

Domain: all real numbers

Range: \((5,\infty)\)

The graph is still increasing because the base is greater than 1. It approaches \(y=5\) from above as \(x\to -\infty\).

Example 4: Describe the transformations of \(y=-3\left(2^{x-1}\right)+4\)

We compare this to the parent function \(y=2^x\).

• \(x-1\) means shift right 1 unit.
• The factor \(-3\) means vertical stretch by factor 3 and reflection in the x-axis.
• The \(+4\) means shift up 4 units.

Horizontal asymptote: The vertical shift gives asymptote

$$y=4$$

y-intercept: Let \(x=0\).

$$y=-3\left(2^{-1}\right)+4=-3\left(\frac12\right)+4=-\frac32+4=\frac52$$

So the y-intercept is

$$\left(0,\frac52\right)$$

Behavior: Because of the reflection, the graph lies below the asymptote for large negative values of the exponent’s effect and falls as \(x\) increases. It approaches \(y=4\) on one side and drops downward on the other side.

7. A Step-by-Step Method for Sketching

When you are given an exponential function, use this method:

1. Identify the parent function, usually \(y=b^x\).
2. Determine whether it is growth or decay by checking the base.
3. Look for transformations: stretches, reflections, and shifts.
4. Find the horizontal asymptote.
5. Find the y-intercept by substituting \(x=0\).
6. Calculate a few extra points if needed.
7. Sketch a smooth curve that approaches the asymptote.

8. Common Mistakes to Avoid

• Forgetting the asymptote changes when the graph shifts up or down. For \(y=2^x-4\), the asymptote is \(y=-4\), not \(y=0\).
• Mixing up horizontal shifts. In \(y=2^{x-3}\), the graph shifts right 3, not left 3.
• Assuming all exponential graphs pass through \((0,1)\). This is only true for the basic form \(y=b^x\). A transformed graph may have a different y-intercept.
• Thinking the graph touches the asymptote. It approaches the asymptote but does not cross it in the usual exponential forms studied here.

9. Summary

Exponential functions have the form \(y=a\,b^x\) or \(y=a\,b^{(x-h)}+k\). Their graphs usually have a horizontal asymptote, a y-intercept, and a smooth curve that shows either growth or decay.

The base tells whether the graph increases or decreases. Transformations such as shifts, reflections, and stretches change the graph’s position and shape. To sketch an exponential function well, always identify the asymptote, the y-intercept, and a few key points.

Logarithms as Inverse Functions

When you see an exponential expression like \(2^5 = 32\), the exponent tells you how many times the base is used as a factor. A logarithm answers the inverse question: what exponent produced this number?

This is why logarithms and exponential functions are called inverse functions. They undo each other, just like addition and subtraction, or squaring and square roots in the right settings.

Understanding this idea is the key to translating between exponential and logarithmic forms, solving equations, and making sense of growth models in higher mathematics.

1. Exponential Form and Logarithmic Form

The statement

$$a^b = c$$

can be rewritten in logarithmic form as

$$\log_a(c) = b$$

These two statements mean exactly the same thing.

• \(a\) is the base
• \(b\) is the exponent
• \(c\) is the result

So if

$$3^4 = 81$$

then the equivalent logarithmic statement is

$$\log_3(81) = 4$$

This reads: “log base 3 of 81 equals 4,” which means “3 raised to what power gives 81?”

2. Why Logarithms Are Inverses

An inverse function reverses the action of another function. If an exponential function raises a base to a power, then a logarithm recovers that power.

For example, let

$$f(x) = 2^x$$

Its inverse is

$$f^{-1}(x) = \log_2(x)$$

These functions undo each other:

$$\log_2(2^x) = x$$

and

$$2^{\log_2(x)} = x$$

as long as the input is valid. Since exponential outputs are always positive, logarithms are only defined for positive inputs.

3. Domain and Range

Because inverse functions swap inputs and outputs, the domain and range also switch.

For the exponential function \(y = a^x\), where \(a > 0\) and \(a \ne 1\):

• Domain: all real numbers
• Range: \((0, \infty)\)

For the logarithmic function \(y = \log_a(x)\):

• Domain: \((0, \infty)\)
• Range: all real numbers

This means:

• You can raise a positive base to any real exponent.
• You can only take the logarithm of a positive number.

4. The General Conversion Rule

To move between forms, remember this pattern:

$$\log_a(c) = b \quad \Longleftrightarrow \quad a^b = c$$

A good way to remember it is:

• The base stays the same.
• The log equals exponent.
• The input of the log becomes the result.

5. Special Logarithms

Two logarithms appear very often in mathematics:

• Common logarithm: \(\log(x)\), which means \(\log_{10}(x)\)
• Natural logarithm: \(\ln(x)\), which means \(\log_e(x)\)

The number \(e\) is a special irrational number approximately equal to \(2.718\). It is important in continuous growth and decay.

Because \(\ln(x)\) is the inverse of \(e^x\), we have:

$$\ln(e^x) = x$$

and

$$e^{\ln(x)} = x$$

for \(x > 0\).

6. Key Inverse Relationships

For any base \(a > 0\), \(a \ne 1\):

$$\log_a(a^x) = x$$

and

$$a^{\log_a(x)} = x$$

These are true because logarithms and exponentials undo each other.

You should also know these basic values:

• \(\log_a(1) = 0\) because \(a^0 = 1\)
• \(\log_a(a) = 1\) because \(a^1 = a\)
• \(\ln(1) = 0\)
• \(\ln(e) = 1\)

7. Worked Examples

Example 1: Rewrite exponential form as logarithmic form

Rewrite \(5^3 = 125\) in logarithmic form.

Step 1: Identify the base, exponent, and result.

• Base: \(5\)
• Exponent: \(3\)
• Result: \(125\)

Step 2: Use the pattern \(a^b = c \iff \log_a(c) = b\).

$$\log_5(125) = 3$$

Answer: \(\log_5(125) = 3\)

Example 2: Rewrite logarithmic form as exponential form

Rewrite \(\log_2(16) = 4\) in exponential form.

Step 1: The base is \(2\), the result is \(16\), and the exponent is \(4\).

Step 2: Write base raised to exponent equals result.

$$2^4 = 16$$

Answer: \(2^4 = 16\)

Example 3: Evaluate a logarithm by thinking of the inverse

Find \(\log_3(\frac{1}{27})\).

Step 1: Ask: “3 raised to what power equals \(\frac{1}{27}\)?”

Since

$$3^3 = 27$$

then

$$3^{-3} = \frac{1}{27}$$

So the exponent is \(-3\).

$$\log_3\left(\frac{1}{27}\right) = -3$$

Answer: \(-3\)

Example 4: Use the natural logarithm as an inverse

Solve \(e^x = 7\).

Step 1: We want to undo the exponential base \(e\).

Step 2: Apply natural logarithm to both sides.

$$\ln(e^x) = \ln(7)$$

Step 3: Use the inverse relationship \(\ln(e^x) = x\).

$$x = \ln(7)$$

Answer: \(x = \ln(7)\)

This answer is exact. As a decimal, \(x \approx 1.946\).

8. Interpreting Logarithms as Questions About Exponents

A logarithm is best understood as a question:

$$\log_a(x) = ?$$

means

“What power must I raise \(a\) to in order to get \(x\)?”

For instance:

• \(\log_2(8) = 3\) because \(2^3 = 8\)
• \(\log_{10}(1000) = 3\) because \(10^3 = 1000\)
• \(\ln(e^5) = 5\) because \(e^5\) already has base \(e\)

This interpretation helps you avoid memorizing too many separate rules. If you think in terms of exponents, logarithms become much easier.

9. Common Mistakes to Avoid

• Mixing up the base and the answer: In \(\log_2(8) = 3\), the answer is not \(8\); the answer is the exponent \(3\).
• Using a negative or zero input: Expressions like \(\log_5(0)\) or \(\ln(-2)\) are not defined in real numbers.
• Forgetting the inverse idea: When solving exponential equations, use logarithms because they undo exponents.
• Confusing \(\log(x)\) and \(\ln(x)\): \(\log(x)\) usually means base 10, while \(\ln(x)\) means base \(e\).

10. Quick Check

1. Rewrite \(4^2 = 16\) in logarithmic form.
2. Rewrite \(\log_7(49) = 2\) in exponential form.
3. Evaluate \(\log_{10}(100)\).
4. Evaluate \(\ln(e^3)\).

Answers:

1. \(\log_4(16) = 2\)
2. \(7^2 = 49\)
3. \(2\)
4. \(3\)

Summary

Logarithms are the inverse of exponential functions. The equation \(a^b = c\) means the same thing as \(\log_a(c) = b\), so a logarithm tells you the exponent needed to produce a number.

This idea allows you to translate between forms, evaluate logarithms, and solve exponential equations. In particular, \(\ln(x)\) and \(e^x\) are inverse functions, which is especially important in continuous growth situations.

Graphs of Logarithmic Functions

Logarithmic functions are the inverse of exponential functions. If an exponential function tells us how a quantity grows or decays, a logarithmic function tells us the exponent needed to produce a certain value.

In this lesson, you will learn how to graph logarithmic functions, identify their domain, range, vertical asymptotes, and x-intercepts, and understand how transformations affect their graphs.

A basic logarithmic function has the form

$$y=\log_b(x)$$

where the base satisfies \(b>0\) and \(b\ne 1\).

This means:

$$b^y=x$$

So the logarithm answers the question: “To what power must we raise \(b\) to get \(x\)?”

For example, if

$$y=\log_2(8)$$

then \(y=3\), because \(2^3=8\).

1. The Basic Shape of a Logarithmic Graph

The parent logarithmic graph is usually written as

$$y=\log_b(x)$$

Its graph has several important features.

• Domain: \(x>0\)
• Range: all real numbers
• Vertical asymptote: \(x=0\)
• x-intercept: \((1,0)\)
• No y-intercept: because \(x=0\) is not in the domain

The reason the domain is only positive numbers is that a logarithm such as \(\log_b(x)\) is only defined when \(x>0\).

For instance, \(\log_{10}(5)\) is defined, but \(\log_{10}(0)\) and \(\log_{10}(-3)\) are not real numbers.

As \(x\) gets closer and closer to \(0\) from the right, the values of \(\log_b(x)\) decrease without bound. That is why the graph approaches the line \(x=0\), but never touches it. This line is called a vertical asymptote.

2. How the Base Affects the Graph

The shape depends on the base \(b\).

Case 1: \(b>1\)

If the base is greater than 1, such as \(2\), \(3\), or \(10\), the graph increases from left to right.

Examples:

$$y=\log_2(x), \quad y=\log_3(x), \quad y=\log_{10}(x)$$

Case 2: \(0<b<1\)

If the base is between 0 and 1, such as \(\frac12\), the graph decreases from left to right.

Example:

$$y=\log_{1/2}(x)$$

Even though the graph decreases, it still has:

• domain \(x>0\)
• range all real numbers
• vertical asymptote \(x=0\)
• x-intercept \((1,0)\)

3. Key Points to Plot

To sketch a logarithmic graph, it helps to use values of \(x\) that are powers of the base.

For the function

$$y=\log_2(x)$$

use the fact that:

• \(\log_2(1)=0\)
• \(\log_2(2)=1\)
• \(\log_2(4)=2\)
• \(\log_2\left(\frac12\right)=-1\)
• \(\log_2\left(\frac14\right)=-2\)

So important points are:

$$ (1,0),\ (2,1),\ (4,2),\ \left(\frac12,-1\right),\ \left(\frac14,-2\right) $$

These points create the familiar logarithmic curve that rises slowly to the right and falls sharply near the asymptote \(x=0\).

4. Relationship to Exponential Graphs

Because logarithmic and exponential functions are inverses, their graphs are reflections of each other across the line

$$y=x$$

For example, the functions

$$y=2^x \quad \text{and} \quad y=\log_2(x)$$

are inverses. If \((a,b)\) is on the graph of \(y=2^x\), then \((b,a)\) is on the graph of \(y=\log_2(x)\).

This helps explain why exponential functions have a horizontal asymptote, while logarithmic functions have a vertical asymptote.

5. Transformations of Logarithmic Graphs

Just like other functions, logarithmic graphs can be shifted, stretched, reflected, and compressed.

A general transformed logarithmic function may look like

$$y=a\log_b(x-h)+k$$

Each part changes the graph in a predictable way.

• \(h\): shifts the graph right if \(h>0\), left if \(h<0\)
• \(k\): shifts the graph up if \(k>0\), down if \(k<0\)
• \(a\): stretches or compresses vertically; if \(a<0\), it also reflects across the x-axis

The vertical asymptote changes with horizontal shifts. For

$$y=\log_b(x-h)$$

the asymptote is

$$x=h$$

The domain must make the logarithm input positive:

$$x-h>0$$

so

$$x>h$$

6. Finding the x-Intercept

The x-intercept occurs where \(y=0\).

For a logarithmic function, set the expression equal to zero and solve.

Example for the parent function:

$$0=\log_b(x)$$

This means

$$x=b^0=1$$

So the x-intercept is always \((1,0)\) for the basic logarithmic function.

For transformed logarithmic functions, the x-intercept may change.

Worked Example 1: Graphing a Basic Logarithmic Function

Graph

$$y=\log_3(x)$$

Step 1: Identify the domain and asymptote.

The input of the logarithm must be positive, so:

$$x>0$$

Therefore, the domain is \((0,\infty)\), and the vertical asymptote is

$$x=0$$

Step 2: Find key points.

• \(\log_3(1)=0\), so \((1,0)\)
• \(\log_3(3)=1\), so \((3,1)\)
• \(\log_3(9)=2\), so \((9,2)\)
• \(\log_3\left(\frac13\right)=-1\), so \(\left(\frac13,-1\right)\)

Step 3: Sketch the curve.

Plot the points and draw a smooth increasing curve that approaches \(x=0\) from the right.

Answer:

• Domain: \(x>0\)
• Range: all real numbers
• Vertical asymptote: \(x=0\)
• x-intercept: \((1,0)\)

Worked Example 2: Identifying Transformations

Graph

$$y=\log_2(x-4)+1$$

Step 1: Compare with the parent function.

The parent is

$$y=\log_2(x)$$

Replacing \(x\) with \(x-4\) shifts the graph 4 units to the right. Adding \(+1\) shifts it up 1 unit.

Step 2: Find the asymptote and domain.

Set the inside equal to 0 for the asymptote:

$$x-4=0$$ $$x=4$$

So the vertical asymptote is \(x=4\).

For the domain:

$$x-4>0$$ $$x>4$$

Step 3: Find some points.

Use values where \(x-4\) is a power of 2.

• If \(x-4=1\), then \(x=5\):

$$y=\log_2(1)+1=0+1=1$$

Point: \((5,1)\)

• If \(x-4=2\), then \(x=6\):

$$y=\log_2(2)+1=1+1=2$$

Point: \((6,2)\)

• If \(x-4=\frac12\), then \(x=4.5\):

$$y=\log_2\left(\frac12\right)+1=-1+1=0$$

Point: \((4.5,0)\)

Step 4: State the x-intercept.

The x-intercept is \((4.5,0)\), or \(\left(\frac92,0\right)\).

Answer:

• Shift right 4, up 1
• Vertical asymptote: \(x=4\)
• Domain: \(x>4\)
• x-intercept: \(\left(\frac92,0\right)\)

Worked Example 3: A Reflected Logarithmic Graph

Graph

$$y=-\log_{10}(x)+2$$

Step 1: Describe the transformations.

The negative sign in front reflects the graph of \(y=\log_{10}(x)\) across the x-axis. Then \(+2\) shifts it up 2 units.

Step 2: Domain and asymptote.

The input is still \(x\), so:

$$x>0$$

The vertical asymptote is still

$$x=0$$

Step 3: Find key points.

• At \(x=1\):

$$y=-\log_{10}(1)+2=-(0)+2=2$$

Point: \((1,2)\)

• At \(x=10\):

$$y=-\log_{10}(10)+2=-1+2=1$$

Point: \((10,1)\)

• At \(x=100\):

$$y=-\log_{10}(100)+2=-2+2=0$$

Point: \((100,0)\)

Step 4: Interpret the graph.

This graph decreases from left to right. It gets very large as \(x\) approaches 0 from the right, and it crosses the x-axis at \((100,0)\).

Worked Example 4: Finding Graph Features from the Equation

Consider

$$y=2\ln(x+3)-4$$

Remember that \(\ln(x)\) means \(\log_e(x)\), where \(e\) is the natural base.

Step 1: Domain and asymptote.

The logarithm input must be positive:

$$x+3>0$$ $$x>-3$$

So the domain is \((-3,\infty)\), and the vertical asymptote is

$$x=-3$$

Step 2: Describe transformations.

• \(x+3\): shift left 3
• Multiply by 2: vertical stretch by factor 2
• \(-4\): shift down 4

Step 3: Find the x-intercept.

Set \(y=0\):

$$0=2\ln(x+3)-4$$ $$4=2\ln(x+3)$$ $$2=\ln(x+3)$$

Rewrite in exponential form:

$$x+3=e^2$$ $$x=e^2-3$$

So the x-intercept is

$$\left(e^2-3,0\right)$$

This is approximately

$$\left(4.39,0\right)$$

7. Steps for Sketching Any Logarithmic Function

1. Find where the logarithm is defined by making its input positive.
2. Use that condition to determine the domain.
3. Find the vertical asymptote by setting the logarithm input equal to 0.
4. Identify transformations from the parent function.
5. Find the x-intercept by setting \(y=0\).
6. Plot a few easy points.
7. Draw a smooth curve that approaches the asymptote but never crosses it.

8. Common Mistakes to Avoid

• Using zero or negative x-values in the domain. Logarithms are only defined for positive inputs.
• Forgetting that the asymptote is vertical. For logarithmic graphs, the asymptote comes from where the inside equals 0.
• Mixing up the intercepts. Most basic logarithmic graphs have an x-intercept, but no y-intercept.
• Ignoring transformations inside the logarithm. A change like \(x-2\) affects the graph horizontally, not vertically.

Brief Summary

A logarithmic function has the form \(y=\log_b(x)\), where \(x>0\). Its graph has a restricted positive domain, a vertical asymptote, and usually an x-intercept.

The parent graph passes through \((1,0)\) and has asymptote \(x=0\). When the function is transformed, the asymptote and x-intercept may move, but the key idea remains the same: the expression inside the logarithm must stay positive.

If you can identify the domain, asymptote, and a few key points, you can successfully sketch almost any logarithmic graph.

Properties of Logarithms help us rewrite logarithmic expressions in simpler forms. These properties are especially useful when we need to expand one logarithm into several terms or condense several logarithms into one term.

In Grade 12, you will often use these rules when solving equations, simplifying expressions, and working with exponential and logarithmic models. Understanding why the rules work and when to use them will make logarithm problems much easier.

Before using logarithm properties, remember what a logarithm means. If

$$\log_b(x)=y,$$

then this means

$$b^y=x,$$

where b b is the base, x b is the argument, and the following must be true:

• \(b>0\)
• \(b\neq 1\)
• \(x>0\)

The most important thing to remember is that logarithms are only defined for positive arguments. So whenever you expand or condense expressions, the quantities inside the logs must stay positive.

The three main properties of logarithms are the product property, quotient property, and power property.

1. Product Property

If \(M>0\) and \(N>0\), then

$$\log_b(MN)=\log_b(M)+\log_b(N).$$

This means that the logarithm of a product becomes a sum of logarithms.

2. Quotient Property

If \(M>0\) and \(N>0\), then

$$\log_b\left(\frac{M}{N}\right)=\log_b(M)-\log_b(N).$$

This means that the logarithm of a quotient becomes a difference of logarithms.

3. Power Property

If \(M>0\) and \(p\) is any real number, then

$$\log_b(M^p)=p\log_b(M).$$

This means that an exponent inside a logarithm can be moved to the front as a coefficient.

These three properties are the main tools for expanding and condensing logarithmic expressions.

Important note: there is no property that allows you to split a sum or difference inside a logarithm. In general,

$$\log_b(M+N)\neq \log_b(M)+\log_b(N)$$

and

$$\log_b(M-N)\neq \log_b(M)-\log_b(N).$$

This is a very common mistake. The product and quotient properties only work for multiplication and division, not addition or subtraction.

How to expand logarithmic expressions

To expand means to rewrite a single logarithm as a sum or difference of simpler logarithms. A good strategy is:

1. Look for multiplication, division, or powers inside the logarithm.
2. Use the product or quotient property first if needed.
3. Use the power property to move exponents to the front.
4. Make sure all arguments remain positive.

Worked Example 1: Expanding a simple product

Expand:

$$\log_2(8x)$$

Since \(8x\) is a product, use the product property:

$$\log_2(8x)=\log_2(8)+\log_2(x).$$

Now simplify \(\log_2(8)\). Since \(2^3=8\),

$$\log_2(8)=3.$$

So the expanded form is

$$\log_2(8x)=3+\log_2(x).$$

This is valid when \(x>0\).

Worked Example 2: Expanding a quotient with a power

Expand:

$$\log\left(\frac{x^3y^2}{5}\right)$$

Here, \(\log\) means base 10. Start with the quotient property:

$$\log\left(\frac{x^3y^2}{5}\right)=\log(x^3y^2)-\log(5).$$

Now use the product property on \(x^3y^2\):

$$\log(x^3y^2)=\log(x^3)+\log(y^2).$$

So we get

$$\log\left(\frac{x^3y^2}{5}\right)=\log(x^3)+\log(y^2)-\log(5).$$

Now use the power property:

$$\log(x^3)=3\log(x), \quad \log(y^2)=2\log(y).$$

Therefore, the fully expanded form is

$$\log\left(\frac{x^3y^2}{5}\right)=3\log(x)+2\log(y)-\log(5).$$

This is valid when \(x>0\) and \(y>0\).

How to condense logarithmic expressions

To condense means to combine several logarithmic terms into a single logarithm. This is the reverse of expanding.

A good strategy is:

1. Use coefficients as exponents by applying the power property in reverse.
2. Combine sums using the product property in reverse.
3. Combine differences using the quotient property in reverse.
4. Write the final answer as one logarithm if possible.

Worked Example 3: Condensing a sum and difference

Condense:

$$2\ln(x)+\ln(y)-\ln(3)$$

Use the power property in reverse on the first term:

$$2\ln(x)=\ln(x^2).$$

Now the expression becomes

$$\ln(x^2)+\ln(y)-\ln(3).$$

Use the product property in reverse to combine the first two terms:

$$\ln(x^2)+\ln(y)=\ln(x^2y).$$

Now use the quotient property in reverse:

$$\ln(x^2y)-\ln(3)=\ln\left(\frac{x^2y}{3}\right).$$

So the condensed form is

$$\ln\left(\frac{x^2y}{3}\right).$$

This is valid when \(x>0\) and \(y>0\).

Worked Example 4: Condensing with multiple coefficients

Condense:

$$\frac{1}{2}\log(a)-3\log(b)+\log(c)$$

First move the coefficients inside as exponents:

$$\frac{1}{2}\log(a)=\log(a^{1/2}), \quad 3\log(b)=\log(b^3).$$

So the expression becomes

$$\log(a^{1/2})-\log(b^3)+\log(c).$$

Combine the sum first:

$$\log(a^{1/2})+\log(c)=\log(a^{1/2}c).$$

Now combine with the difference:

$$\log(a^{1/2}c)-\log(b^3)=\log\left(\frac{a^{1/2}c}{b^3}\right).$$

So the condensed form is

$$\log\left(\frac{\sqrt{a}c}{b^3}\right).$$

Since \(a^{1/2}=\sqrt{a}\), both forms are acceptable.

Using the natural logarithm

The natural logarithm, written as \(\ln(x)\), is simply a logarithm with base \(e\), where \(e\approx 2.718\). All the same properties apply:

• $$\ln(MN)=\ln(M)+\ln(N)$$
• $$\ln\left(\frac{M}{N}\right)=\ln(M)-\ln(N)$$
• $$\ln(M^p)=p\ln(M)$$

These rules are especially important in exponential growth and continuous growth problems because equations often involve \(e\) and \(\ln\).

Common mistakes to avoid

• Do not split sums or differences inside a logarithm.
  For example, \(\ln(x+2)\) cannot be rewritten as \(\ln(x)+\ln(2)\).
• Do not forget domain restrictions.
  If you write \(\log(x)\) or \(\ln(x)\), then \(x\) must be positive.
• Be careful with coefficients.
  A coefficient in front of a logarithm becomes an exponent on the argument, not a multiplication inside the log.
• Keep track of signs.
  A subtraction between logs means division when condensing.

Quick reference table

• Product: $$\log_b(MN)=\log_b(M)+\log_b(N)$$
• Quotient: $$\log_b\left(\frac{M}{N}\right)=\log_b(M)-\log_b(N)$$
• Power: $$\log_b(M^p)=p\log_b(M)$$

When expanding, go from one logarithm to many.

When condensing, go from many logarithms to one.

Brief Summary

The properties of logarithms let you rewrite expressions in useful ways. The product property changes multiplication into addition, the quotient property changes division into subtraction, and the power property moves exponents to the front. These rules work for all valid logarithm bases, including base 10 and base \(e\), and they are essential for simplifying expressions and solving logarithmic equations.

Change of Base Formula is a logarithm rule that lets us rewrite a logarithm in one base using a different base. This is especially useful because calculators usually have buttons for only common logarithms \,\(\log x\,\) (base 10) and natural logarithms \,\(\ln x\,\) (base \(e\)).

For example, most calculators do not have a button for \,\(\log_3 20\,\). The change of base formula allows us to rewrite it using logs we can calculate directly.

This lesson will explain what the formula means, why it works, how to use it, and how it helps when solving logarithmic and exponential problems.

1. Review: What does a logarithm mean?

A logarithm answers the question: “To what power must the base be raised to get the number?”

For instance, if

$$3^x = 81$$

then

$$x = \log_3 81$$

Since \,\(3^4 = 81\,\), we know that

$$\log_3 81 = 4$$

So logarithms and exponents are inverse operations.

2. The Change of Base Formula

The formula is:

$$\log_b a = \frac{\log a}{\log b}$$

This version uses common logarithms, or base 10 logs.

You can also use natural logarithms:

$$\log_b a = \frac{\ln a}{\ln b}$$

Both forms give the same answer. You can use whichever is more convenient on your calculator.

Important conditions:

• \(a > 0\)
• \(b > 0\)
• \(b \ne 1\)

These are the usual conditions for logarithms to be defined.

3. Why the formula works

Suppose

$$x = \log_b a$$

By the meaning of a logarithm, this means

$$b^x = a$$

Now take the logarithm of both sides. Using common logs, we get

$$\log(b^x) = \log a$$

Use the power rule \,\(\log(m^n) = n\log m\,\):

$$x\log b = \log a$$

Now divide both sides by \,\(\log b\,\):

$$x = \frac{\log a}{\log b}$$

Since \,\(x = \log_b a\,\), we have

$$\log_b a = \frac{\log a}{\log b}$$

The exact same reasoning works with natural logs:

$$\log_b a = \frac{\ln a}{\ln b}$$

4. When do we use the change of base formula?

You use it when:

• the logarithm has a base other than 10 or \(e\),
• you want a decimal approximation,
• you are solving an exponential equation and need to isolate a variable,
• you want to compare logarithms written in different bases.

5. Worked Example 1: Evaluating a logarithm with an unusual base

Evaluate \,\(\log_2 7\,\) to three decimal places.

Step 1: Apply the change of base formula.

$$\log_2 7 = \frac{\log 7}{\log 2}$$

Step 2: Use a calculator.

$$\log 7 \approx 0.8451, \qquad \log 2 \approx 0.3010$$ $$\log_2 7 \approx \frac{0.8451}{0.3010} \approx 2.807$$

Answer: \,\(\log_2 7 \approx 2.807\,\)

This makes sense because \,\(2^2 = 4\,\) and \,\(2^3 = 8\,\), so the exponent needed to get 7 should be between 2 and 3.

6. Worked Example 2: Using natural logs instead

Evaluate \,\(\log_5 18\,\) to three decimal places using natural logs.

Step 1: Rewrite using the formula.

$$\log_5 18 = \frac{\ln 18}{\ln 5}$$

Step 2: Calculate.

$$\ln 18 \approx 2.8904, \qquad \ln 5 \approx 1.6094$$ $$\log_5 18 \approx \frac{2.8904}{1.6094} \approx 1.796$$

Answer: \,\(\log_5 18 \approx 1.796\,\)

Again, this is reasonable because \,\(5^1 = 5\,\) and \,\(5^2 = 25\,\), and 18 is between 5 and 25.

7. Worked Example 3: Solving an exponential equation

Solve

$$3^x = 20$$

Step 1: Rewrite in logarithmic form.

$$x = \log_3 20$$

Step 2: Use the change of base formula.

$$x = \frac{\log 20}{\log 3}$$

Step 3: Approximate.

$$\log 20 \approx 1.3010, \qquad \log 3 \approx 0.4771$$ $$x \approx \frac{1.3010}{0.4771} \approx 2.727$$

Answer: \,\(x \approx 2.727\,\)

This means \,\(3^{2.727}\,\) is approximately 20.

8. Worked Example 4: A continuous growth context with base \(e\)

Suppose an amount grows continuously according to

$$A = 500e^{0.08t}$$

How long will it take for the amount to reach 900?

Step 1: Substitute \,\(A = 900\,\).

$$900 = 500e^{0.08t}$$

Step 2: Isolate the exponential expression.

$$\frac{900}{500} = e^{0.08t}$$ $$1.8 = e^{0.08t}$$

Step 3: Take the natural log of both sides.

$$\ln(1.8) = \ln(e^{0.08t})$$

Since \,\(\ln(e^x) = x\,\), this becomes

$$\ln(1.8) = 0.08t$$

Step 4: Solve for \,\(t\,\).

$$t = \frac{\ln(1.8)}{0.08}$$ $$t \approx \frac{0.5878}{0.08} \approx 7.35$$

Answer: It will take about \,\(7.35\,\) time units.

In continuous growth problems, natural logs are especially useful because the model uses the base \,\(e\,\).

9. Common mistakes to avoid

• Mixing different log types in one fraction: Do not write \,\(\frac{\log a}{\ln b}\,\). Use either both common logs or both natural logs.
• Reversing the numerator and denominator: \,\(\log_b a = \frac{\log a}{\log b}\,\), not \,\(\frac{\log b}{\log a}\,\).
• Forgetting calculator parentheses: Type the entire numerator and denominator carefully, such as \,\((\ln 18)/(\ln 5)\,\).
• Thinking the base changes the answer: The rewritten form looks different, but the value stays the same.

10. Helpful observations

If the base and argument are powers of the same number, you may not need the change of base formula.

For example,

$$\log_4 64 = 3$$

because \,\(4^3 = 64\,\).

But if the relationship is not obvious, the change of base formula is the best tool.

You should also know that the formula can use any valid base \,\(k\,\):

$$\log_b a = \frac{\log_k a}{\log_k b}$$

In practice, we usually choose \,\(k=10\,\) or \,\(k=e\,\) because calculators provide those directly.

11. Quick practice ideas

• Rewrite \,\(\log_7 15\,\) using common logs.
• Rewrite \,\(\log_4 11\,\) using natural logs.
• Solve \,\(2^x = 13\,\).
• Solve \,\(7^x = 100\,\).

12. Summary

The change of base formula lets us rewrite logarithms in a base we can compute with a calculator.

$$\log_b a = \frac{\log a}{\log b} = \frac{\ln a}{\ln b}$$

This is useful for evaluating logarithms with unusual bases and for solving exponential equations. Remember to keep the same type of logarithm in both the numerator and denominator, and be careful with the order: argument on top, base on bottom.

Solving Exponential Equations

Exponential equations are equations in which the variable appears in the exponent. For example, in the equation \(2^x = 16\), the variable \(x\) is in the exponent.

These equations are important because they are used to model real situations such as population growth, radioactive decay, compound interest, and continuous growth with the base \(e\).

To solve exponential equations, we usually use one of three main ideas:

• Rewrite both sides with the same base.
• Use the inverse relationship between exponentials and logarithms.
• Take the logarithm of both sides when the equation cannot be rewritten with a common base.

Let’s build these methods step by step.

1. Solving by rewriting with a common base

If both sides of the equation can be written using the same base, then we can set the exponents equal.

This works because if \(a^m = a^n\) and \(a > 0\), \(a \ne 1\), then \(m = n\).

For example, suppose we want to solve:

$$2^{x+1} = 8$$

Rewrite \(8\) as a power of 2:

$$8 = 2^3$$

So the equation becomes:

$$2^{x+1} = 2^3$$

Now the bases are the same, so set the exponents equal:

$$x+1=3$$ $$x=2$$

This is often the fastest method when it works.

2. Solving using logarithms

Sometimes the two sides cannot be rewritten with the same base. In that case, we use logarithms.

Remember the inverse relationship:

$$a^x = b \iff x = \log_a b$$

If the equation is not easy to rewrite, take a logarithm of both sides. You may use common logarithm \(\log\) or natural logarithm \(\ln\). Both will give the same answer if used correctly.

For example, solve:

$$3^x = 20$$

This cannot be rewritten neatly with a common base, so take \(\log\) of both sides:

$$\log(3^x) = \log(20)$$

Use the power rule of logarithms, \(\log(a^b)=b\log(a)\):

$$x\log(3) = \log(20)$$

Now solve for \(x\):

$$x = \frac{\log(20)}{\log(3)}$$

Using a calculator,

$$x \approx 2.727$$

3. Solving equations with the natural base \(e\)

In many Grade 12 applications, especially continuous growth and decay, the base is \(e\). The natural logarithm \(\ln\) is especially useful here because \(\ln\) and \(e^x\) are inverse functions.

The key inverse property is:

$$\ln(e^x)=x \quad \text{and} \quad e^{\ln x}=x$$

For example, solve:

$$e^{2x} = 7$$

Take \(\ln\) of both sides:

$$\ln(e^{2x}) = \ln(7)$$

Using the inverse property,

$$2x = \ln(7)$$ $$x = \frac{\ln(7)}{2}$$

4. Important logarithm facts to use

When solving exponential equations, these logarithm rules are especially helpful:

• \(\log(a^b) = b\log(a)\)
• \(\ln(a^b) = b\ln(a)\)
• \(\ln(e^x)=x\)
• If \(a^m = a^n\), then \(m=n\) when \(a>0\) and \(a\ne1\)

Be careful: there is no rule that says \(\log(a+b)=\log a + \log b\). That is not true.

Worked Example 1: Common base method

Solve:

$$5^{2x} = 125$$

Step 1: Rewrite \(125\) as a power of 5.

$$125 = 5^3$$

Step 2: Set the exponents equal.

$$5^{2x} = 5^3 \Rightarrow 2x=3$$

Step 3: Solve.

$$x=\frac{3}{2}$$

Answer: \(x=\frac{3}{2}\)

Worked Example 2: Variable on both sides

Solve:

$$4^x = 2^{x+3}$$

Step 1: Rewrite both sides using base 2.

$$4^x = (2^2)^x = 2^{2x}$$

So the equation becomes:

$$2^{2x} = 2^{x+3}$$

Step 2: Set exponents equal.

$$2x = x+3$$ $$x=3$$

Answer: \(x=3\)

Worked Example 3: Use logarithms

Solve:

$$7^{x-1} = 12$$

This cannot be rewritten easily with a common base, so take logarithms of both sides:

$$\log(7^{x-1}) = \log(12)$$

Apply the power rule:

$$ (x-1)\log(7) = \log(12) $$

Now solve for \(x\):

$$x-1 = \frac{\log(12)}{\log(7)}$$ $$x = 1 + \frac{\log(12)}{\log(7)}$$

Approximate value:

$$x \approx 2.277$$

Answer: \(x = 1 + \frac{\log(12)}{\log(7)} \approx 2.277\)

Worked Example 4: Natural base in a continuous context

A quantity grows according to the equation

$$A = 500e^{0.3t}$$

Find \(t\) when \(A=1000\).

Step 1: Substitute \(A=1000\).

$$1000 = 500e^{0.3t}$$

Step 2: Divide both sides by 500.

$$2 = e^{0.3t}$$

Step 3: Take \(\ln\) of both sides.

$$\ln(2) = \ln(e^{0.3t})$$ $$\ln(2) = 0.3t$$

Step 4: Solve for \(t\).

$$t = \frac{\ln(2)}{0.3}$$ $$t \approx 2.31$$

Answer: \(t \approx 2.31\)

A step-by-step strategy

1. Look at the equation and ask: Can both sides be written with the same base?
2. If yes, rewrite and set the exponents equal.
3. If no, isolate the exponential expression if needed.
4. Take \(\log\) or \(\ln\) of both sides.
5. Use the power rule to bring the exponent down.
6. Solve the resulting linear equation.
7. Check your answer in the original equation if possible.

Common mistakes to avoid

• Not rewriting carefully: For example, \(8=2^3\), not \(2^4\).
• Forgetting the power rule: \(\log(3^x)=x\log(3)\).
• Using the wrong inverse idea: \(\ln(e^{2x})=2x\), not just \(x\).
• Rounding too early: Keep calculator values exact until the final step when possible.
• Dropping brackets: \(\log(7^{x-1})=(x-1)\log 7\), so the whole \((x-1)\) must stay together.

When should you use \(\log\) and when should you use \(\ln\)?

Either can be used for most exponential equations. However:

• Use \(\ln\) especially when the base is \(e\).
• Use \(\log\) or \(\ln\) when the base is something else, depending on your preference or calculator.

For example, to solve \(5^x=9\), both of these are correct:

$$x = \frac{\log 9}{\log 5}$$

or

$$x = \frac{\ln 9}{\ln 5}$$

Brief Summary

To solve exponential equations, first try rewriting both sides with a common base. If that is not possible, use logarithms to bring the exponent down where you can solve for the variable. In equations involving \(e\), the natural logarithm \(\ln\) is especially useful because it is the inverse of the exponential function with base \(e\).

With practice, the key is choosing the right method: common base if possible, logarithms if not.

Solving Logarithmic Equations means finding the value of a variable when that variable appears inside a logarithm.

This topic is closely connected to exponential equations because logarithms and exponentials are inverse operations. In many problems, you solve by rewriting a logarithmic equation in exponential form or by combining logarithms first and then exponentiating.

One of the most important ideas in this lesson is that not every algebraic answer is valid. Because logarithms are only defined for positive inputs, you must always check that every logarithmic argument is greater than 0. This is why logarithmic equations often produce extraneous solutions, which are answers that appear during solving but do not actually work in the original equation.

By the end of this lesson, you should be able to:

• identify when a logarithmic equation can be rewritten in exponential form,
• use logarithm properties to combine or simplify expressions,
• solve equations involving common logs, natural logs, and other bases,
• check whether a solution is valid by testing the logarithmic arguments.

1. Review: What a logarithm means

A logarithm answers the question: “To what power must the base be raised to get this number?”

For example,

$$\log_2 8 = 3$$

because

$$2^3 = 8$$

In general,

$$\log_b A = c \iff b^c = A$$

This equivalence is the main tool for solving logarithmic equations.

2. Domain rule: logarithmic arguments must be positive

Before solving, always remember:

$$\log_b(M) \text{ is defined only when } M > 0$$

So if you see an equation like

$$\log(x-4) = 2$$

then the argument must satisfy

$$x - 4 > 0 \quad \Rightarrow \quad x > 4$$

This condition helps you reject impossible answers later.

3. Main strategies for solving logarithmic equations

1. If there is a single logarithm by itself, rewrite the equation in exponential form.
2. If there are multiple logarithms, use logarithm properties to combine them into one logarithm if possible.
3. After solving, check every answer in the original equation and make sure each logarithmic argument is positive.

4. Important logarithm properties

These properties are often used to combine logarithms:

• Product rule: $$\log_b M + \log_b N = \log_b(MN)$$
• Quotient rule: $$\log_b M - \log_b N = \log_b\left(\frac{M}{N}\right)$$
• Power rule: $$\log_b(M^k) = k\log_b M$$

These rules only work when the logarithms have the same base and when all arguments are positive.

5. Solving by rewriting in exponential form

If you have just one logarithm, isolate it and then convert to exponential form.

For example, if

$$\ln(x+1) = 2$$

then rewrite using base \(e\):

$$x+1 = e^2$$

Then solve:

$$x = e^2 - 1$$

Since \(x+1 = e^2 > 0\), the answer is valid.

Worked Example 1: Basic logarithmic equation

Solve:

$$\log_3(x-1) = 2$$

Step 1: Check the domain.

$$x - 1 > 0 \quad \Rightarrow \quad x > 1$$

Step 2: Rewrite in exponential form.

$$x - 1 = 3^2$$

Step 3: Solve.

$$x - 1 = 9$$ $$x = 10$$

Step 4: Check.

The argument is \(10 - 1 = 9\), which is positive, and

$$\log_3 9 = 2$$

So the solution is

$$\boxed{x = 10}$$

6. Solving equations with natural logarithms

The natural logarithm, written \(\ln x\), is a logarithm with base \(e\).

That means:

$$\ln A = c \iff e^c = A$$

This is especially useful in continuous growth and decay contexts.

Worked Example 2: Natural logarithm equation

Solve:

$$\ln(2x-5) = 1$$

Step 1: Domain.

$$2x - 5 > 0 \quad \Rightarrow \quad x > \frac{5}{2}$$

Step 2: Rewrite in exponential form.

$$2x - 5 = e^1$$ $$2x - 5 = e$$

Step 3: Solve.

$$2x = e + 5$$ $$x = \frac{e+5}{2}$$

Step 4: Check.

Substituting gives

$$2\left(\frac{e+5}{2}\right) - 5 = e > 0$$

and

$$\ln(e) = 1$$

So the solution is

$$\boxed{x = \frac{e+5}{2}}$$

7. Solving equations with more than one logarithm

When an equation contains two or more logarithms, combine them into a single logarithm whenever possible.

Then use the fact that if

$$\log_b A = \log_b B$$

then

$$A = B$$

as long as both arguments are positive.

Worked Example 3: Combining logarithms

Solve:

$$\log(x) + \log(x-9) = 1$$

Here, \(\log\) means base 10.

Step 1: Domain.

$$x > 0 \quad \text{and} \quad x-9 > 0$$ $$x > 9$$

Step 2: Use the product rule.

$$\log(x(x-9)) = 1$$

Step 3: Rewrite in exponential form.

$$x(x-9) = 10^1$$ $$x(x-9) = 10$$

Step 4: Solve the quadratic equation.

$$x^2 - 9x - 10 = 0$$ $$ (x-10)(x+1)=0$$

So the possible solutions are

$$x=10 \quad \text{or} \quad x=-1$$

Step 5: Check against the domain.

The domain requires \(x > 9\). So:

• \(x=10\) is valid
• \(x=-1\) is invalid because logarithms of negative numbers are undefined

Therefore, the only solution is

$$\boxed{x=10}$$

This example shows why checking is essential. The algebra produced two answers, but only one works in the original logarithmic equation.

8. Equations that create extraneous solutions

Extraneous solutions often appear when solving logarithmic equations that lead to quadratics or rational expressions.

You should always test answers in the original equation, not just in a simplified version.

Worked Example 4: A logarithmic equation with an extraneous solution

Solve:

$$\ln(x+2) - \ln(x-1) = \ln 4$$

Step 1: Domain.

$$x+2 > 0 \quad \text{and} \quad x-1 > 0$$

The stricter condition is

$$x > 1$$

Step 2: Use the quotient rule.

$$\ln\left(\frac{x+2}{x-1}\right)=\ln 4$$

Step 3: Set arguments equal.

$$\frac{x+2}{x-1}=4$$

Step 4: Solve.

$$x+2 = 4(x-1)$$ $$x+2 = 4x-4$$ $$6 = 3x$$ $$x=2$$

Step 5: Check.

Since \(2 > 1\), both logarithmic arguments are positive:

$$2+2=4, \quad 2-1=1$$

Now test in the original equation:

$$\ln 4 - \ln 1 = \ln 4 - 0 = \ln 4$$

So the solution is

$$\boxed{x=2}$$

9. Special note: when both sides are logarithms

If the equation is already in the form

$$\log_b(f(x)) = \log_b(g(x))$$

then you can set the arguments equal:

$$f(x)=g(x)$$

But do not forget the domain conditions:

$$f(x) > 0 \quad \text{and} \quad g(x) > 0$$

For example, if

$$\ln(x+3)=\ln(2x-1)$$

then

$$x+3=2x-1$$ $$x=4$$

Check: \(x+3=7>0\) and \(2x-1=7>0\), so it works.

10. Common mistakes to avoid

• Forgetting the domain. Every logarithmic argument must be positive.
• Not checking solutions. An answer from algebra may not work in the original equation.
• Using log rules incorrectly. For example, \(\log(a+b)\neq \log a + \log b\).
• Mixing bases. You can only combine logarithms directly when they have the same base.
• Dropping parentheses. In \(\log(x-2)\), the entire expression \(x-2\) is the argument.

11. Step-by-step method you can always use

1. Identify all logarithmic arguments and write any domain restrictions.
2. Isolate logarithms if needed.
3. Use logarithm properties to combine into one logarithm if possible.
4. Rewrite in exponential form, or set arguments equal if both sides are logs with the same base.
5. Solve the resulting equation.
6. Check each solution in the original equation.

12. Quick practice thinking

Ask yourself these questions when solving:

• Are all logarithms the same base?
• Can I combine them using product or quotient rules?
• Do I need to exponentiate to remove the logarithm?
• Does my final answer make every log argument positive?

Summary

To solve logarithmic equations, use the inverse relationship between logarithms and exponentials. If there is one logarithm, rewrite in exponential form. If there are several logarithms, combine them first using log properties whenever possible.

Most importantly, always check your answers in the original equation. A value is only a true solution if every logarithmic argument is positive and the equation is satisfied.

Logistic Growth Models are used when a quantity grows quickly at first, but then the growth slows down as it approaches a maximum possible value.

This is different from ordinary exponential growth. In exponential growth, the quantity keeps increasing faster and faster without any limit. In many real-life situations, that is not realistic.

For example, a population of bacteria in a dish cannot grow forever because food, space, and nutrients are limited. A town population cannot grow without limit if housing and resources are limited. In these cases, a logistic model is more appropriate.

The main idea is that growth is fast when the population is small, but the growth rate decreases as the population gets closer to a maximum value called the carrying capacity.

The logistic growth function is usually written as

$$P(t)=\frac{K}{1+A e^{-rt}}$$

where:

• \(P(t)\) is the population or quantity at time \(t\),
• \(K\) is the carrying capacity, the greatest value the population approaches,
• \(A\) is a constant determined by the starting value,
• \(r\) is the growth rate,
• \(e\) is the natural base.

This formula creates an S-shaped curve, also called a sigmoid curve.

Why the graph is S-shaped

At the beginning, when \(t\) is small, the denominator is not too large, so the population can rise quickly.

In the middle stage, the population is growing rapidly, often close to its fastest rate of increase.

Later, as \(P(t)\) gets closer to \(K\), growth slows down because the model is being limited by the carrying capacity.

Eventually, the graph levels off and approaches \(K\), but in the model it does not usually exceed it.

Important features of a logistic model

• The population stays positive.
• The population approaches \(K\) as \(t\to\infty\).
• The graph has a horizontal asymptote at \(y=K\).
• The fastest growth occurs at the middle level \(P=\frac{K}{2}\).

That last fact is very important. In a logistic model, the population grows fastest when it reaches half of the carrying capacity.

Finding the constant \(A\)

If the initial population is known, we can find \(A\). Suppose \(P(0)=P_0\). Substitute \(t=0\) into the logistic formula:

$$P(0)=\frac{K}{1+A e^{0}}=\frac{K}{1+A}$$

So if \(P(0)=P_0\), then

$$P_0=\frac{K}{1+A}$$

Solve for \(A\):

$$1+A=\frac{K}{P_0}$$ $$A=\frac{K}{P_0}-1=\frac{K-P_0}{P_0}$$

This lets us build a logistic equation from the carrying capacity, initial value, and growth rate.

Comparing exponential and logistic growth

• Exponential growth: unlimited increase, model often looks like \(P(t)=P_0e^{rt}\).
• Logistic growth: growth slows over time because of a maximum limit, model looks like \(P(t)=\frac{K}{1+A e^{-rt}}\).

If a problem mentions limited resources, crowding, or a maximum sustainable population, logistic growth is usually the better model.

Worked Example 1: Evaluating a logistic function

A fish population in a lake is modeled by

$$P(t)=\frac{5000}{1+9e^{-0.4t}}$$

where \(t\) is measured in years. Find:

1. the carrying capacity,
2. the initial population,
3. the population after 5 years.

Step 1: Identify the carrying capacity

The carrying capacity is the number on top of the fraction:

$$K=5000$$

So the population approaches 5000 fish.

Step 2: Find the initial population

Substitute \(t=0\):

$$P(0)=\frac{5000}{1+9e^0}=\frac{5000}{1+9}$$ $$P(0)=\frac{5000}{10}=500$$

So the initial population is 500 fish.

Step 3: Find the population after 5 years

Substitute \(t=5\):

$$P(5)=\frac{5000}{1+9e^{-0.4(5)}}=\frac{5000}{1+9e^{-2}}$$

Using \(e^{-2}\approx 0.1353\):

$$P(5)\approx\frac{5000}{1+9(0.1353)}=\frac{5000}{1+1.2177}=\frac{5000}{2.2177}$$ $$P(5)\approx 2255.5$$

So after 5 years, the population is about 2256 fish.

Worked Example 2: Building a logistic model from given information

A rabbit population has a carrying capacity of 1200. Initially there are 150 rabbits, and the growth rate constant is \(r=0.6\). Write the logistic model.

Step 1: Start with the logistic form

$$P(t)=\frac{K}{1+A e^{-rt}}$$

Substitute the known values \(K=1200\) and \(r=0.6\):

$$P(t)=\frac{1200}{1+A e^{-0.6t}}$$

Step 2: Use the initial value to find \(A\)

We know \(P(0)=150\), so

$$150=\frac{1200}{1+A}$$

Solve:

$$1+A=\frac{1200}{150}=8$$ $$A=7$$

Final model:

$$P(t)=\frac{1200}{1+7e^{-0.6t}}$$

This equation describes the rabbit population over time.

Worked Example 3: Solving for time using logarithms

A population is modeled by

$$P(t)=\frac{800}{1+3e^{-0.5t}}$$

When will the population reach 600?

Step 1: Set \(P(t)=600\)

$$600=\frac{800}{1+3e^{-0.5t}}$$

Step 2: Solve for the exponential expression

$$600(1+3e^{-0.5t})=800$$ $$1+3e^{-0.5t}=\frac{800}{600}=\frac{4}{3}$$ $$3e^{-0.5t}=\frac{4}{3}-1=\frac{1}{3}$$ $$e^{-0.5t}=\frac{1}{9}$$

Step 3: Use natural logarithms

$$\ln(e^{-0.5t})=\ln\left(\frac{1}{9}\right)$$ $$-0.5t=\ln\left(\frac{1}{9}\right)$$

Since \(\ln(1/9)=-\ln 9\),

$$t=\frac{-\ln(1/9)}{0.5}=\frac{\ln 9}{0.5}$$

Using \(\ln 9\approx 2.197\):

$$t\approx\frac{2.197}{0.5}=4.394$$

So the population reaches 600 at about 4.39 time units.

Worked Example 4: Finding when growth is fastest

A population is modeled by

$$P(t)=\frac{2000}{1+4e^{-0.3t}}$$

At what population level is the growth fastest, and when does that happen?

Step 1: Use the logistic model property

In a logistic model, growth is fastest when the population is half the carrying capacity.

Since \(K=2000\), the fastest growth occurs at

$$P=\frac{2000}{2}=1000$$

Step 2: Solve for the time when \(P(t)=1000\)

$$1000=\frac{2000}{1+4e^{-0.3t}}$$ $$1000(1+4e^{-0.3t})=2000$$ $$1+4e^{-0.3t}=2$$ $$4e^{-0.3t}=1$$ $$e^{-0.3t}=\frac{1}{4}$$

Take natural logs:

$$-0.3t=\ln\left(\frac{1}{4}\right)$$ $$t=\frac{-\ln(1/4)}{0.3}=\frac{\ln 4}{0.3}$$

Using \(\ln 4\approx 1.386\):

$$t\approx\frac{1.386}{0.3}=4.62$$

So the growth is fastest when the population is 1000, which happens at about 4.62 time units.

How to recognize a logistic model in a word problem

Look for phrases like:

• approaches a maximum,
• limited by resources,
• carrying capacity,
• grows quickly at first and then slows,
• S-shaped growth curve.

These clues suggest that the quantity does not grow without bound and should be modeled logistically.

Common mistakes to avoid

• Confusing \(K\) with the initial value: \(K\) is the maximum possible population, not the starting population.
• Forgetting that \(e^0=1\): this is needed when finding \(P(0)\).
• Using the wrong sign in the exponent: logistic models are usually written with \(e^{-rt}\).
• Not using logarithms correctly: when solving for time, isolate the exponential first, then apply \(\ln\).
• Assuming the graph is exponential: logistic growth levels off, while exponential growth keeps rising without a horizontal limit.

Quick review of solving logistic equations

1. Substitute the known value of \(P(t)\).
2. Clear the fraction by multiplying both sides.
3. Isolate the exponential term.
4. Take the natural logarithm of both sides.
5. Solve for \(t\).

Summary

A logistic growth model describes growth that starts quickly, then slows as it approaches a maximum carrying capacity.

The standard form is

$$P(t)=\frac{K}{1+A e^{-rt}}$$

Here, \(K\) is the carrying capacity, and the graph has an S-shape with horizontal asymptote \(y=K\).

Logistic models are useful for populations and other real situations where growth is limited by resources. They can be analyzed by finding initial values, evaluating the function, solving for time with logarithms, and identifying that the fastest growth happens at \(P=\frac{K}{2}\).