Exponential Functions and Models

Laws of Exponents Revisited

Exponents appear whenever a quantity is multiplied by itself repeatedly. In Algebra 2 and 11th Grade Maths, exponent rules are especially important because they let us simplify complicated expressions quickly and accurately.

When students make mistakes with exponent laws, it is often because they remember the rule but forget when it applies. This lesson reviews the main laws of exponents and shows how to use them correctly in more complex expressions with multiple variables.

By the end of this lesson, you should be able to simplify expressions involving product rules, quotient rules, powers of powers, negative exponents, and zero exponents.

1. What an exponent means

If \(a\) is a number and \(n\) is a positive integer, then \(a^n\) means multiplying \(a\) by itself \(n\) times:

$$a^n = \underbrace{a \cdot a \cdot a \cdots a}_{n \text{ factors}}$$

For example, \(x^4 = x \cdot x \cdot x \cdot x\).

This idea is the foundation of every exponent rule. The laws are not random shortcuts—they come from repeated multiplication.

2. The main laws of exponents

Here are the exponent laws you must know.

• Product of powers: when multiplying same bases, add exponents.

$$a^m \cdot a^n = a^{m+n}$$

• Quotient of powers: when dividing same bases, subtract exponents.

$$\frac{a^m}{a^n} = a^{m-n} \quad (a \ne 0)$$

• Power of a power: when raising a power to another power, multiply exponents.

$$\left(a^m\right)^n = a^{mn}$$

• Power of a product: distribute the exponent to each factor.

$$\left(ab\right)^n = a^n b^n$$

• Power of a quotient: distribute the exponent to the numerator and denominator.

$$\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \quad (b \ne 0)$$

• Zero exponent: any nonzero base raised to the zero power equals 1.

$$a^0 = 1 \quad (a \ne 0)$$

• Negative exponent: a negative exponent means reciprocal.

$$a^{-n} = \frac{1}{a^n} \quad (a \ne 0)$$

3. A very important warning

Exponent rules work only in certain situations. The most common mistake is trying to combine terms that do not have the same base or that are being added instead of multiplied.

For example:

• \(x^2 \cdot x^5 = x^7\) because the bases are the same and you are multiplying.
• \(x^2 + x^5\) cannot be simplified using exponent laws because this is addition, not multiplication.
• \(x^2 \cdot y^2\) cannot become \((xy)^4\). The bases are different.

Always ask yourself: Am I multiplying or dividing powers with the same base?

4. Product rule and quotient rule with variables

When variables match, combine their exponents. When they do not match, leave them separate.

Examples:

• \(a^3 \cdot a^4 = a^7\)
• \(m^8 \div m^3 = m^5\)
• \(x^2 y^5 \cdot x^3 y = x^{2+3} y^{5+1} = x^5 y^6\)
• \(\frac{p^7 q^2}{p^3 q} = p^{7-3} q^{2-1} = p^4 q\)

Notice that each variable follows its own exponent rule. You can think of each variable as having its own “track.”

5. Power rules

The power of a power rule tells us to multiply exponents:

$$\left(x^3\right)^4 = x^{12}$$

This works because

$$\left(x^3\right)^4 = x^3 \cdot x^3 \cdot x^3 \cdot x^3 = x^{3+3+3+3} = x^{12}$$

If a power is applied to a product or quotient, distribute it to every factor inside the parentheses.

Examples:

• \((2x^3)^2 = 2^2(x^3)^2 = 4x^6\)
• \((ab^2)^3 = a^3 b^6\)
• \(\left(\frac{3x^2}{y}\right)^2 = \frac{9x^4}{y^2}\)

6. Zero exponents and why they make sense

The rule \(a^0 = 1\) may seem strange at first, but it follows from the quotient rule.

For example,

$$\frac{x^5}{x^5} = x^{5-5} = x^0$$

But any nonzero quantity divided by itself is 1, so

$$x^0 = 1 \quad (x \ne 0)$$

This is why expressions like \((3ab)^0\) simplify to 1, as long as the base is not zero.

7. Negative exponents

A negative exponent does not mean the value is negative. It means the factor belongs on the other side of the fraction bar.

Examples:

• \(x^{-3} = \frac{1}{x^3}\)
• \(\frac{1}{y^{-2}} = y^2\)
• \(a^{-2}b^3 = \frac{b^3}{a^2}\)

A final answer is usually written with positive exponents only.

8. Strategy for simplifying complex exponent expressions

When an expression looks complicated, do not try to do everything at once. Use a step-by-step process.

1. Simplify powers first, especially powers of powers.
2. Distribute exponents across products or quotients inside parentheses.
3. Combine like bases by adding or subtracting exponents.
4. Rewrite negative exponents as reciprocals.
5. Check that the final answer has no negative exponents and is fully simplified.

9. Worked Examples

Example 1: Simplify a product with several variables

Simplify:

$$3x^2y^3 \cdot 2x^5y^4$$

Step 1: Multiply the coefficients.

$$3 \cdot 2 = 6$$

Step 2: Add exponents for matching bases.

$$x^2 \cdot x^5 = x^7$$ $$y^3 \cdot y^4 = y^7$$

Answer:

$$3x^2y^3 \cdot 2x^5y^4 = 6x^7y^7$$

This example uses the product rule.

Example 2: Simplify a quotient with negative exponents in the result

Simplify:

$$\frac{12a^7b^2}{3a^3b^5}$$

Step 1: Simplify the coefficients.

$$\frac{12}{3} = 4$$

Step 2: Subtract exponents for matching bases.

$$a^{7-3} = a^4$$ $$b^{2-5} = b^{-3}$$

So the expression becomes

$$4a^4b^{-3}$$

Step 3: Rewrite with positive exponents.

$$4a^4b^{-3} = \frac{4a^4}{b^3}$$

Answer:

$$\frac{12a^7b^2}{3a^3b^5} = \frac{4a^4}{b^3}$$

Example 3: Simplify a power of a product and quotient

Simplify:

$$\left(\frac{2x^3y^{-2}}{z^2}\right)^2$$

Step 1: Apply the exponent to every factor.

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

Step 2: Multiply exponents.

$$= \frac{4x^6y^{-4}}{z^4}$$

Step 3: Rewrite the negative exponent.

$$\frac{4x^6y^{-4}}{z^4} = \frac{4x^6}{y^4z^4}$$

Answer:

$$\left(\frac{2x^3y^{-2}}{z^2}\right)^2 = \frac{4x^6}{y^4z^4}$$

Example 4: Simplify a more complex expression

Simplify:

$$\frac{(3m^2n^{-1})^3(2m^{-4}n^5)}{6m^3n^2}$$

Step 1: Simplify the power first.

$$ (3m^2n^{-1})^3 = 3^3(m^2)^3(n^{-1})^3 = 27m^6n^{-3} $$

Now substitute this back in:

$$\frac{27m^6n^{-3}(2m^{-4}n^5)}{6m^3n^2}$$

Step 2: Multiply in the numerator.

$$27 \cdot 2 = 54$$ $$m^6 \cdot m^{-4} = m^2$$ $$n^{-3} \cdot n^5 = n^2$$

So the expression becomes

$$\frac{54m^2n^2}{6m^3n^2}$$

Step 3: Simplify the quotient.

$$\frac{54}{6} = 9$$ $$m^{2-3} = m^{-1}$$ $$n^{2-2} = n^0 = 1$$

So we get

$$9m^{-1}$$

Step 4: Rewrite with positive exponents.

$$9m^{-1} = \frac{9}{m}$$

Answer:

$$\frac{(3m^2n^{-1})^3(2m^{-4}n^5)}{6m^3n^2} = \frac{9}{m}$$

10. Common mistakes to avoid

• Adding exponents when bases are different: \(x^2y^3\) cannot become \((xy)^5\).
• Using exponent rules across addition: \(x^2 + x^3\) does not equal \(x^5\).
• Forgetting to distribute an outer exponent to every factor: \((2xy)^3 = 8x^3y^3\), not \(2x^3y^3\).
• Leaving negative exponents in the final answer: rewrite them as reciprocals.
• Forgetting zero exponent rule: \(a^0 = 1\) for any nonzero \(a\).

11. Quick check for understanding

Try these on your own:

1. Simplify \(x^4 \cdot x^{-2}\)
2. Simplify \(\frac{p^5q^3}{p^2q^6}\)
3. Simplify \((2a^{-1}b^2)^3\)
4. Simplify \(\frac{(m^2n)^2}{mn^3}\)

Answers:

1. \(x^2\)
2. \(\frac{p^3}{q^3}\)
3. \(\frac{8b^6}{a^3}\)
4. \(m^3n^{-1} = \frac{m^3}{n}\)

12. Summary

The laws of exponents help you simplify expressions by using patterns in repeated multiplication. Multiply same bases by adding exponents, divide same bases by subtracting exponents, and multiply exponents when taking a power of a power.

Negative exponents mean reciprocals, and zero exponents equal 1 as long as the base is nonzero. In complex multivariable expressions, work carefully step by step, combine only matching bases, and write final answers with positive exponents.

Defining the Exponential Function

In many real-world situations, a quantity does not change by adding the same amount each time. Instead, it changes by multiplying by the same factor over equal intervals. This kind of change is called exponential change.

An exponential function is usually written in the form

$$f(x)=ab^x$$

where:

• (a\) is the initial value, or the value of the function when \(x=0\),
• \(b\) is the base, or the constant factor by which the function changes,
• \(x\) is the input, often representing time or number of steps.

Understanding this form helps you recognize whether a situation shows growth or decay, and helps you graph the function correctly.

1. What makes a function exponential?

A function is exponential when the variable is in the exponent. In other words, the input \(x\) appears as a power.

For example:

• \(f(x)=3(2)^x\) is exponential because \(x\) is in the exponent.
• \(g(x)=5x^2\) is not exponential because \(x\) is the base being squared.
• \(h(x)=7\left(\frac{1}{2}\right)^x\) is exponential.

This is an important difference. In exponential functions, the output changes by multiplying by the same factor as \(x\) increases by 1.

2. The meaning of \(a\) and \(b\) in \(f(x)=ab^x\)

Let us look more closely at the two constants in the formula.

The value of \(a\)

The number \(a\) tells you the starting amount. This is because when \(x=0\),

$$f(0)=ab^0=a(1)=a$$

So the y-intercept of the graph is \((0,a)\).

The value of \(b\)

The number \(b\) tells you the factor the function is multiplied by each time \(x\) increases by 1.

For example, if \(b=3\), then each step multiplies the output by 3. If \(b=\frac{1}{2}\), then each step multiplies the output by \(\frac{1}{2}\).

You can see this by comparing consecutive outputs:

$$f(x+1)=ab^{x+1}=ab^x\cdot b=f(x)\cdot b$$

So every time \(x\) goes up by 1, the output is multiplied by \(b\).

3. Conditions on the base

For \(f(x)=ab^x\) to be an exponential function in the usual algebra sense, the base must satisfy:

• \(b>0\)
• \(b\neq 1\)

These conditions matter because:

• If \(b\leq 0\), the function is not defined for all real values of \(x\).
• If \(b=1\), then \(f(x)=a\cdot 1^x=a\), which is just a constant function, not exponential growth or decay.

In many school examples, \(a>0\) as well, especially when modeling populations, money, or measurements.

4. Exponential growth vs. exponential decay

The value of the base \(b\) tells you whether the function shows growth or decay.

Exponential growth

If

$$b>1$$

then the function increases as \(x\) increases. This is called exponential growth.

Examples:

• \(f(x)=2(3)^x\)
• \(g(x)=100(1.05)^x\)

In \(100(1.05)^x\), the factor \(1.05\) means the quantity grows by 5% each step.

Exponential decay

If

$$0<b<1$$

then the function decreases as \(x\) increases. This is called exponential decay.

Examples:

• \(f(x)=5\left(\frac{1}{2}\right)^x\)
• \(g(x)=80(0.9)^x\)

In \(80(0.9)^x\), the factor \(0.9\) means the quantity keeps 90% of its value each step, so it decreases by 10% each step.

5. Key features of the graph

Graphs of exponential functions have some important features you should know.

• Y-intercept: The graph crosses the y-axis at \((0,a)\).
• Growth or decay: Determined by whether \(b>1\) or \(0<b<1\).
• Horizontal asymptote: For basic functions \(f(x)=ab^x\), the graph gets closer and closer to \(y=0\), but does not touch it.
• Domain: All real numbers.
• Range: If \(a>0\), then \(f(x)>0\), so the range is \((0,\infty)\).

If \(a>0\), the graph stays above the x-axis. As \(x\) gets very large, a growth function rises quickly, while a decay function gets closer and closer to 0.

6. Recognizing exponential patterns in tables

A table represents an exponential function when equal changes in \(x\) produce equal multiplicative changes in the output.

For example, consider this table:

$$\begin{array}{c|c} x & f(x) \\\hline 0 & 4 \\ 1 & 8 \\ 2 & 16 \\ 3 & 32 \end{array}$$

Each output is multiplied by 2, so this is exponential. The function is

$$f(x)=4(2)^x$$

Compare that with a linear pattern:

$$\begin{array}{c|c} x & f(x) \\\hline 0 & 4 \\ 1 & 7 \\ 2 & 10 \\ 3 & 13 \end{array}$$

Here, the outputs increase by adding 3 each time. That is linear, not exponential.

7. Worked Examples

Example 1: Identify whether the function is exponential growth or decay

Classify the function:

$$f(x)=6\left(\frac{1}{3}\right)^x$$

Step 1: Check the form.

This matches \(f(x)=ab^x\), where \(a=6\) and \(b=\frac{1}{3}\).

Step 2: Look at the base.

Since \(0<\frac{1}{3}<1\), the function shows exponential decay.

Step 3: Find the initial value.

When \(x=0\),

$$f(0)=6\left(\frac{1}{3}\right)^0=6$$

Answer: This is an exponential decay function with initial value 6.

Example 2: Find the equation from a table

A function has the following values:

$$\begin{array}{c|c} x & f(x) \\\hline 0 & 5 \\ 1 & 15 \\ 2 & 45 \\ 3 & 135 \end{array}$$

Write the exponential function.

Step 1: Find \(a\).

Since \(f(0)=a\), we get

$$a=5$$

Step 2: Find the common factor.

Check the ratio of consecutive outputs:

$$\frac{15}{5}=3, \quad \frac{45}{15}=3, \quad \frac{135}{45}=3$$

So the base is

$$b=3$$

Step 3: Write the function.

$$f(x)=5(3)^x$$

Answer: The exponential function is \(f(x)=5(3)^x\).

Example 3: Evaluate and interpret an exponential function

Suppose

$$P(x)=200(1.5)^x$$

a) What is the initial value?

b) Is this growth or decay?

c) Find \(P(2)\).

Step 1: Identify \(a\) and \(b\).

Here, \(a=200\) and \(b=1.5\).

a) The initial value is

$$P(0)=200$$

b) Since \(1.5>1\), this is exponential growth.

c) Evaluate at \(x=2\):

$$P(2)=200(1.5)^2=200(2.25)=450$$

Answer: The initial value is 200, the function represents growth, and \(P(2)=450\).

Example 4: Decide if a function is exponential

Which of the following are exponential functions?

1. \(f(x)=4^x\)
2. \(g(x)=x^4\)
3. \(h(x)=7(0.8)^x\)
4. \(k(x)=3x+2\)

Step 1: Check whether the variable is in the exponent.

• \(f(x)=4^x\): yes, exponential
• \(g(x)=x^4\): no, not exponential
• \(h(x)=7(0.8)^x\): yes, exponential
• \(k(x)=3x+2\): no, linear

Answer: The exponential functions are \(f(x)=4^x\) and \(h(x)=7(0.8)^x\).

8. Common mistakes to avoid

• Confusing exponential and polynomial functions: \(x^2\) is not exponential, but \(2^x\) is.
• Forgetting the base rules: The base must be positive and not equal to 1.
• Mixing up growth and decay: If \(b>1\), it is growth. If \(0<b<1\), it is decay.
• Using addition instead of multiplication: Exponential functions change by multiplying by the same factor, not adding the same amount.

9. How to quickly identify an exponential function

When you see a function, table, or graph, ask these questions:

1. Is the variable in the exponent?
2. Can it be written as \(ab^x\)?
3. Is the output changing by a constant ratio?
4. Is the base greater than 1 or between 0 and 1?

If the answer to these fits the pattern of \(ab^x\), then you are working with an exponential function.

Summary

An exponential function has the form $$f(x)=ab^x$$ where \(a\) is the initial value and \(b\) is the constant multiplicative factor. If \(b>1\), the function represents exponential growth. If \(0<b<1\), it represents exponential decay.

To recognize an exponential function, look for the variable in the exponent and check whether outputs change by a constant ratio. The graph crosses the y-axis at \((0,a)\) and, for basic exponential functions, has a horizontal asymptote at \(y=0\).

Introduction

In exponential functions, the base tells us how a quantity grows or changes. You may already know bases like 2, 3, or 10. Another very important base is a special number called e.

The number e appears naturally when growth happens in smaller and smaller time intervals. It is the base used for continuous growth, such as certain population models, interest earned continuously, and many real-world situations in science and finance.

In this lesson, we will learn where e comes from, how it is connected to the expression $$\left(1+\frac{1}{n}\right)^n,$$ and why it is so useful in exponential models.

1. A special limit

Consider the expression

$$\left(1+\frac{1}{n}\right)^n.$$

As the value of \(n\) gets larger and larger, this expression gets closer and closer to a fixed number. That number is called \(e\).

So we define

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

This means that when \(n\) becomes extremely large, the value of $$\left(1+\frac{1}{n}\right)^n$$ approaches \(e\).

The decimal value of \(e\) is approximately

$$e\approx 2.71828.$$

Like \(\pi\), this decimal goes on forever without repeating.

2. Why does this expression matter?

This expression comes from repeated growth. Suppose you start with 1 unit of money and it grows by 100% in one year.

If that growth happens all at once at the end of the year, you get

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

If it is split into 2 equal growth periods of 50% each, you get

$$\left(1+\frac{1}{2}\right)^2=1.5^2=2.25.$$

If it is split into 4 equal growth periods of 25% each, you get

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

If it is split into 12 equal periods, you get

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

As the number of growth periods keeps increasing, the final amount gets closer to \(2.71828\ldots\), not infinitely large. That limiting value is \(e\).

This is why \(e\) is connected to continuous compounding and continuous growth.

3. Looking at the pattern numerically

Let us calculate a few values of $$\left(1+\frac{1}{n}\right)^n$$ to see the pattern.

• When \(n=1\): $$\left(1+1\right)^1=2$$
• When \(n=2\): $$\left(1+\frac{1}{2}\right)^2=2.25$$
• When \(n=5\): $$\left(1+\frac{1}{5}\right)^5\approx 2.48832$$
• When \(n=10\): $$\left(1+\frac{1}{10}\right)^{10}\approx 2.59374$$
• When \(n=100\): $$\left(1+\frac{1}{100}\right)^{100}\approx 2.70481$$
• When \(n=1000\): $$\left(1+\frac{1}{1000}\right)^{1000}\approx 2.71692$$

These values keep getting closer to \(2.71828\ldots\), which shows the approach toward \(e\).

4. What makes \(e\) special?

The number \(e\) is the base for the exponential function

$$y=e^x.$$

This function is especially useful because it models situations where the rate of growth depends on the current amount. In simple terms, the more there is, the faster it grows.

Examples include:

• population growth under ideal conditions,
• money growing with continuous compounding,
• radioactive decay and other natural decay models.

In 11th Grade maths, the main idea is that \(e\) is the natural base for continuous change.

5. Continuous growth formula

For repeated compounding, you may have seen a formula like

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

where:

• \(P\) is the starting amount,
• \(r\) is the annual rate written as a decimal,
• \(n\) is the number of compounding periods per year,
• \(t\) is the number of years,
• \(A\) is the final amount.

When compounding happens continuously, this formula becomes

$$A=Pe^{rt}.$$

This is one of the most important uses of \(e\).

6. Worked Example 1: Approximating \(e\)

Use the expression $$\left(1+\frac{1}{n}\right)^n$$ with \(n=20\) to estimate \(e\).

Step 1: Substitute \(n=20\).

$$\left(1+\frac{1}{20}\right)^{20}=\left(1.05\right)^{20}.$$

Step 2: Evaluate.

$$1.05^{20}\approx 2.6533.$$

Answer: This gives an estimate of \(e\) as about \(2.6533\).

This is not exact, but it is reasonably close. If we used a larger value of \(n\), the estimate would be better.

7. Worked Example 2: Comparing larger values of \(n\)

Which is a better approximation of \(e\):

$$\left(1+\frac{1}{10}\right)^{10} \quad \text{or} \quad \left(1+\frac{1}{100}\right)^{100}?$$

Step 1: Evaluate both.

$$\left(1+\frac{1}{10}\right)^{10}=1.1^{10}\approx 2.59374$$ $$\left(1+\frac{1}{100}\right)^{100}=1.01^{100}\approx 2.70481$$

Step 2: Compare each to \(e\approx 2.71828\).

• \(2.59374\) is farther from \(2.71828\)
• \(2.70481\) is closer to \(2.71828\)

Answer: $$\left(1+\frac{1}{100}\right)^{100}$$ is the better approximation.

This makes sense because larger values of \(n\) make the expression closer to \(e\).

8. Worked Example 3: Continuous growth

An amount of money grows continuously at a rate of 6% per year. If the initial amount is \(\$500\), how much will there be after 3 years?

Use the continuous growth formula

$$A=Pe^{rt}.$$

Step 1: Identify the values.

• \(P=500\)
• \(r=0.06\)
• \(t=3\)

Step 2: Substitute.

$$A=500e^{0.06(3)}=500e^{0.18}.$$

Step 3: Evaluate.

$$e^{0.18}\approx 1.1972$$ $$A\approx 500(1.1972)=598.60$$

Answer: After 3 years, the amount is approximately \(\$598.60\).

9. Worked Example 4: Comparing yearly compounding and continuous compounding

Suppose \(\$1000\) is invested at 8% interest for 1 year.

Yearly compounding:

$$A=1000(1+0.08)=1080.$$

Continuous compounding:

$$A=1000e^{0.08}.$$

Now evaluate:

$$e^{0.08}\approx 1.08329$$ $$A\approx 1000(1.08329)=1083.29$$

Answer:

• Yearly compounding gives \(\$1080.00\)
• Continuous compounding gives about \(\$1083.29\)

Continuous compounding gives a slightly larger amount because growth is being added constantly.

10. Important ideas to remember

• \(e\) is a special irrational number with decimal value approximately \(2.71828\).
• It is defined by the limit $$e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n.$$
• The expression $$\left(1+\frac{1}{n}\right)^n$$ gets closer to \(e\) as \(n\) gets larger.
• \(e\) is the natural base for continuous growth.
• The continuous growth formula is $$A=Pe^{rt}.$$

11. Common mistakes

• Thinking \(e\) is a variable: \(e\) is a constant number, like \(\pi\).
• Rounding too early: Keep more decimal places in calculations to get a more accurate answer.
• Using percent instead of decimal: In formulas, 6% must be written as \(0.06\), not 6.
• Confusing repeated compounding with continuous compounding: Use $$A=P\left(1+\frac{r}{n}\right)^{nt}$$ for compounding \(n\) times per year, and $$A=Pe^{rt}$$ for continuous compounding.

Summary

The natural base \(e\) is a special number that appears when we study the expression $$\left(1+\frac{1}{n}\right)^n$$ as \(n\) becomes very large. Its value is approximately \(2.71828\).

It is called the base for continuous growth because it is used in models where change happens constantly. In these situations, the main formula is $$A=Pe^{rt}.$$ Understanding \(e\) helps you connect exponential functions to real-life growth and decay.

Solving Exponential Equations with Common Bases

Exponential equations are equations where the variable appears in the exponent. In this lesson, you will learn how to solve a special and very useful type of exponential equation: equations that can be rewritten so that both sides have the same base.

This method is based on a simple idea. If two powers have the same base and are equal, then their exponents must be equal.

For example, if

$$2^5 = 2^5$$

then the exponents are the same. More generally, if

$$a^m = a^n$$

and the base satisfies \(a > 0\) and \(a \ne 1\), then

$$m = n$$

This is the key rule we use to solve exponential equations with common bases.

Why this works

Exponential functions with base greater than 0 and not equal to 1 are one-to-one. That means each output comes from only one exponent value. So if two powers with the same valid base are equal, the exponents must match.

Main strategy

1. Rewrite each side of the equation using the same base, if possible.

2. Once the bases match, set the exponents equal.

3. Solve the resulting linear or simple algebraic equation.

4. Check your answer in the original equation.

Recognizing common bases

Often, numbers can be rewritten as powers of smaller numbers. Here are some useful examples:

• \(4 = 2^2\)

• \(8 = 2^3\)

• \(16 = 2^4\)

• \(25 = 5^2\)

• \(27 = 3^3\)

• \(32 = 2^5\)

• \(81 = 3^4\)

• \(125 = 5^3\)

Being able to spot these relationships helps you rewrite the equation with matching bases.

Exponent rules you will use

When rewriting expressions, these exponent rules are especially helpful:

• Power of a power: \((a^m)^n = a^{mn}\)

• Same base when multiplying: \(a^m \cdot a^n = a^{m+n}\)

• Same base when dividing: \(\dfrac{a^m}{a^n} = a^{m-n}\), for \(a \ne 0\)

Worked Example 1: A direct common-base equation

Solve:

$$2^{x+1} = 2^5$$

Step 1: Check the bases.

Both sides already have base \(2\).

Step 2: Set the exponents equal.

$$x + 1 = 5$$

Step 3: Solve.

$$x = 4$$

Step 4: Check.

$$2^{4+1} = 2^5$$ $$2^5 = 2^5$$

The solution is correct.

Answer: \(x = 4\)

Worked Example 2: Rewrite one side using a common base

Solve:

$$4^x = 2^6$$

Step 1: Rewrite \(4\) as a power of \(2\).

Since \(4 = 2^2\), we have

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

Step 2: Use the power rule.

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

So the equation becomes

$$2^{2x} = 2^6$$

Step 3: Set the exponents equal.

$$2x = 6$$

Step 4: Solve.

$$x = 3$$

Check.

$$4^3 = 64 \quad \text{and} \quad 2^6 = 64$$

The solution works.

Answer: \(x = 3\)

Worked Example 3: Rewrite both sides

Solve:

$$8^{x-1} = 4^{x+2}$$

Step 1: Rewrite both bases as powers of \(2\).

Since \(8 = 2^3\) and \(4 = 2^2\), we get

$$ (2^3)^{x-1} = (2^2)^{x+2} $$

Step 2: Apply the power rule.

$$2^{3(x-1)} = 2^{2(x+2)}$$

Step 3: Set the exponents equal.

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

Step 4: Solve the equation.

$$3x - 3 = 2x + 4$$ $$x = 7$$

Check.

Left side:

$$8^{7-1} = 8^6$$

Right side:

$$4^{7+2} = 4^9$$

Rewrite both:

$$8^6 = (2^3)^6 = 2^{18}$$ $$4^9 = (2^2)^9 = 2^{18}$$

Both sides are equal, so the solution is correct.

Answer: \(x = 7\)

Worked Example 4: Use exponent rules before matching bases

Solve:

$$\frac{3^{x+4}}{3^2} = 27$$

Step 1: Simplify the left side using the quotient rule.

$$\frac{3^{x+4}}{3^2} = 3^{x+4-2} = 3^{x+2}$$

So the equation becomes

$$3^{x+2} = 27$$

Step 2: Rewrite \(27\) as a power of \(3\).

$$27 = 3^3$$

Now we have

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

Step 3: Set the exponents equal.

$$x + 2 = 3$$

Step 4: Solve.

$$x = 1$$

Check.

$$\frac{3^{1+4}}{3^2} = \frac{3^5}{3^2} = 3^3 = 27$$

The solution is correct.

Answer: \(x = 1\)

What to do step by step

• Look at the numbers in the equation and ask: can they be written using the same base?

• Rewrite numbers like \(4, 8, 9, 16, 25, 27, 32, 81, 125\) as powers of smaller numbers.

• Use exponent rules carefully.

• Only set exponents equal after the bases are identical.

• Always check your final answer in the original equation.

Common mistakes to avoid

• Setting exponents equal too early. You can only do this when the bases are the same.

• Forgetting to rewrite numbers. For example, in \(4^x = 2^6\), you must rewrite \(4\) as \(2^2\).

• Using exponent rules incorrectly. For example, \((2^2)^x = 2^{2x}\), not \(2^{x+2}\).

• Not distributing properly. For example, \(3(x-1) = 3x-3\), not \(3x-1\).

• Skipping the check. A quick check can catch algebra mistakes.

Practice problems

Try these on your own:

1. \(5^{x-2} = 5^4\)

2. \(9^x = 3^8\)

3. \(16^{x+1} = 2^{12}\)

4. \(27^x = 9^{x+1}\)

Answers

1. \(x = 6\)

2. \(x = 4\)

3. \(x = 2\)

4. \(x = 2\)

Brief Summary

To solve exponential equations with common bases, rewrite both sides so they use the same base. Then set the exponents equal and solve the resulting equation. This method works because powers with the same valid base are equal only when their exponents are equal.

Compound Interest is a financial application of exponential growth. It describes what happens when interest is added to an amount of money, and then future interest is calculated on both the original amount and the interest already earned.

This is different from simple interest, where interest is only calculated on the original principal. With compound interest, your money can grow faster because the growth builds on itself over time.

In this lesson, you will learn how to use the discrete compound interest formula, how compounding periods affect growth, and how to solve common problems involving future value and interest earned.

1. The basic idea of compound interest

Suppose you invest money in a bank account. The starting amount is called the principal. If the account earns interest, the bank adds a percentage of the balance to your account.

With compound interest, once interest is added, the new balance becomes the amount used to calculate the next interest payment. This repeated multiplication creates exponential growth.

2. The compound interest formula

The formula for compound interest is

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

where:

• \(A\) = final amount after interest
• \(P\) = principal, or starting amount
• \(r\) = annual interest rate written as a decimal
• \(n\) = number of times interest is compounded per year
• \(t\) = time in years

Important: The interest rate must be written as a decimal, not a percent. For example:

• \(5\% = 0.05\)
• \(3.2\% = 0.032\)
• \(12\% = 0.12\)

3. Understanding the parts of the formula

The expression \(1 + \frac{r}{n}\) is the growth factor for one compounding period. It tells you how much the balance is multiplied by each time interest is added.

The exponent \(nt\) tells you the total number of compounding periods. For example:

• If interest is compounded yearly, then \(n = 1\)
• If compounded quarterly, then \(n = 4\)
• If compounded monthly, then \(n = 12\)
• If compounded daily, then \(n = 365\) in many problems

So if money is compounded monthly for 3 years, the total number of compounding periods is

$$nt = 12 \cdot 3 = 36$$

4. Why compound interest is exponential

Each time interest is added, the balance is multiplied by the same factor. Repeated multiplication is the key idea of an exponential function.

For example, if an amount grows by a factor of \(1.02\) each period, then after many periods the amount is found by multiplying by \(1.02\) again and again. That is why exponents appear in compound interest formulas.

5. Common compounding periods

• Annually: \(n = 1\)
• Semiannually: \(n = 2\)
• Quarterly: \(n = 4\)
• Monthly: \(n = 12\)
• Weekly: \(n = 52\)
• Daily: \(n = 365\)

The more often interest is compounded, the more money the account earns, assuming the same annual rate and time.

6. Worked Example 1: Annual compounding

Problem: Find the value of \(\$2000\) invested at \(6\%\) annual interest for 4 years, compounded annually.

Step 1: Identify the values.

• \(P = 2000\)
• \(r = 0.06\)
• \(n = 1\)
• \(t = 4\)

Step 2: Substitute into the formula.

$$A = 2000\left(1 + \frac{0.06}{1}\right)^{1\cdot 4}$$ $$A = 2000(1.06)^4$$

Step 3: Calculate.

$$A = 2000(1.26247696)$$ $$A \approx 2524.95$$

Answer: The investment will be worth approximately \(\$2524.95\) after 4 years.

7. Worked Example 2: Quarterly compounding

Problem: Find the value of \(\$5000\) invested at \(8\%\) annual interest for 3 years, compounded quarterly.

Step 1: Identify the values.

• \(P = 5000\)
• \(r = 0.08\)
• \(n = 4\)
• \(t = 3\)

Step 2: Substitute.

$$A = 5000\left(1 + \frac{0.08}{4}\right)^{4\cdot 3}$$ $$A = 5000(1.02)^{12}$$

Step 3: Calculate.

$$A = 5000(1.26824179)$$ $$A \approx 6341.21$$

Answer: The investment will be worth approximately \(\$6341.21\).

Interest earned is the final amount minus the principal:

$$6341.21 - 5000 = 1341.21$$

So the account earned \(\$1341.21\) in interest.

8. Worked Example 3: Monthly compounding

Problem: A savings account starts with \(\$1200\) and earns \(5.4\%\) annual interest compounded monthly. How much is in the account after 2 years?

Step 1: Identify the values.

• \(P = 1200\)
• \(r = 0.054\)
• \(n = 12\)
• \(t = 2\)

Step 2: Substitute into the formula.

$$A = 1200\left(1 + \frac{0.054}{12}\right)^{12\cdot 2}$$ $$A = 1200(1.0045)^{24}$$

Step 3: Calculate.

$$A \approx 1200(1.113755)$$ $$A \approx 1336.51$$

Answer: After 2 years, the account balance is approximately \(\$1336.51\).

9. Worked Example 4: Comparing compounding periods

Problem: Suppose \(\$10{,}000\) is invested at \(7\%\) annual interest for 1 year. Compare annual compounding and monthly compounding.

Annual compounding:

$$A = 10000\left(1 + \frac{0.07}{1}\right)^{1\cdot 1}$$ $$A = 10000(1.07) = 10700$$

Monthly compounding:

$$A = 10000\left(1 + \frac{0.07}{12}\right)^{12}$$ $$A = 10000(1.0058333\ldots)^{12}$$ $$A \approx 10722.90$$

Comparison:

• Annual compounding gives \(\$10{,}700.00\)
• Monthly compounding gives about \(\$10{,}722.90\)

Monthly compounding earns a little more because interest is added more often.

10. Finding interest earned

Sometimes a problem asks for the interest earned, not the final amount.

Use this relationship:

$$\text{Interest} = A - P$$

So first find the final amount using the compound interest formula, then subtract the original principal.

11. How to solve compound interest problems step by step

1. Read the problem carefully.
2. Identify the principal \(P\).
3. Convert the percent rate to a decimal for \(r\).
4. Determine the compounding frequency \(n\).
5. Identify the time in years \(t\).
6. Substitute into $$A = P\left(1 + \frac{r}{n}\right)^{nt}$$
7. Use a calculator carefully, especially with parentheses and exponents.
8. If needed, subtract \(P\) from \(A\) to find interest earned.
9. Round money to the nearest cent unless told otherwise.

12. Common mistakes to avoid

• Forgetting to change percent to decimal: \(6\%\) should be \(0.06\), not \(6\).
• Using the wrong value of \(n\): Monthly means \(12\), quarterly means \(4\).
• Forgetting the exponent is \(nt\): Multiply the number of periods per year by the number of years.
• Not using parentheses correctly: The entire quantity \(1 + \frac{r}{n}\) must be raised to the exponent.
• Giving interest earned when the question asks for final amount, or vice versa: Read carefully.

13. Connecting compound interest to exponential functions

The compound interest formula is an example of an exponential model because the variable \(t\) appears in the exponent.

If the principal and interest rate stay fixed, the amount grows exponentially over time. This means growth starts gradually and then becomes faster as the balance increases.

That is why compound interest is often described as “interest earning interest.”

14. Quick practice questions

• What is the value of \(\$3000\) at \(4\%\) annual interest for 5 years compounded annually?
• What is the value of \(\$1500\) at \(6\%\) annual interest for 2 years compounded monthly?
• If \(\$8000\) grows to \(\$9261\), how much interest was earned?
• Which earns more over the same time: quarterly compounding or monthly compounding?

15. Brief summary

Compound interest happens when interest is added to an account and future interest is calculated on the new balance. The formula is

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

To use it correctly, identify the principal, convert the rate to a decimal, use the correct compounding number, and multiply \(n\) and \(t\) for the exponent. More frequent compounding leads to greater growth, and the amount grows exponentially over time.

Continuous Growth and Decay Models describe situations where a quantity changes at a rate that is always proportional to its current amount. This means the larger the amount is, the faster it grows or decays, and the smaller it is, the slower it changes.

These models are used in many real-life situations, such as continuously compounded interest, population growth, and radioactive decay. The main formula you will use is

$$A = Pe^{rt}$$

In this formula:

• A is the amount after time \(t\),
• \(P\) is the initial amount,
• \(e\) is a special irrational number approximately equal to \(2.718\),
• \(r\) is the growth or decay rate written as a decimal,
• \(t\) is the time elapsed.

If \(r > 0\), the model represents continuous growth. If \(r < 0\), the model represents continuous decay.

For example, if money grows continuously at a rate of 4%, then \(r = 0.04\). If a substance decays continuously at a rate of 12%, then \(r = -0.12\).

Why do we use \(e\)? In continuous growth and decay, the change is happening all the time, not just once per year or once per month. The number \(e\) appears naturally when compounding happens continuously.

It may help to compare this with ordinary exponential models like \(A = P(1+r)^t\). In that model, growth happens in steps. In the continuous model \(A = Pe^{rt}\), growth happens smoothly at every moment.

Main ideas to remember:

• Use \(A = Pe^{rt}\) when the problem says continuous growth, continuous decay, or compounded continuously.
• The rate \(r\) must be written as a decimal.
• Time \(t\) must match the units of the rate.
• Decay means the exponent is negative because \(r\) is negative.

Understanding the variables carefully is very important. If the rate is given per year, then time should be measured in years. If the rate is given per month, then time should be measured in months.

For instance, if a population grows continuously at 3% per year for 5 years, then

$$A = Pe^{0.03(5)}$$

But if the same 3% rate were somehow given per month, then the time would need to be in months instead.

Continuous growth happens when the amount increases over time. Since \(r\) is positive, the exponent \(rt\) becomes positive, so \(e^{rt}\) is greater than 1. That makes the final amount bigger than the initial amount.

Continuous decay happens when the amount decreases over time. Since \(r\) is negative, the exponent \(rt\) becomes negative, so \(e^{rt}\) is between 0 and 1. That makes the final amount smaller than the initial amount.

Let us now work through several examples.

Example 1: Continuously compounded interest

A bank account starts with \(\$2000\) and earns 5% interest compounded continuously. How much money will be in the account after 3 years?

Step 1: Identify the values.

• \(P = 2000\)
• \(r = 0.05\)
• \(t = 3\)

Step 2: Substitute into the formula.

$$A = Pe^{rt}$$

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

$$A = 2000e^{0.15}$$

Step 3: Evaluate.

Using a calculator, \(e^{0.15} \approx 1.1618\).

$$A \approx 2000(1.1618)$$

$$A \approx 2323.67$$

Answer: After 3 years, the account will have about \(\$2323.67\).

This is a growth problem because the rate is positive.

Example 2: Population growth

A town has a population of 12,000 people. The population grows continuously at a rate of 2.4% per year. What will the population be after 10 years?

Step 1: Identify the values.

• \(P = 12000\)
• \(r = 0.024\)
• \(t = 10\)

Step 2: Use the formula.

$$A = 12000e^{0.024(10)}$$

$$A = 12000e^{0.24}$$

Step 3: Calculate.

\(e^{0.24} \approx 1.2712\)

$$A \approx 12000(1.2712)$$

$$A \approx 15254.4$$

Answer: The population after 10 years will be about 15,254 people.

Since population is usually counted in whole people, rounding to the nearest whole number makes sense.

Example 3: Radioactive decay

A sample begins with 80 grams of a radioactive substance. It decays continuously at a rate of 18% per year. How much remains after 6 years?

Step 1: Identify the values.

• \(P = 80\)
• \(r = -0.18\) because it is decay
• \(t = 6\)

Step 2: Substitute.

$$A = 80e^{-0.18(6)}$$

$$A = 80e^{-1.08}$$

Step 3: Calculate.

\(e^{-1.08} \approx 0.3396\)

$$A \approx 80(0.3396)$$

$$A \approx 27.17$$

Answer: About 27.17 grams remain after 6 years.

Notice that the amount became smaller because the exponent was negative.

Example 4: Half-life and finding the decay rate

A certain substance has a half-life of 8 years. That means half of the original amount remains after 8 years. If the initial mass is 500 grams, how much remains after 20 years?

For half-life problems, we often first find the decay rate \(r\).

Step 1: Use the meaning of half-life.

After 8 years, the amount is half of the initial amount:

$$\frac{1}{2}P = Pe^{8r}$$

Divide both sides by \(P\):

$$\frac{1}{2} = e^{8r}$$

To solve for \(r\), take the natural logarithm of both sides:

$$\ln\left(\frac{1}{2}\right) = 8r$$

$$r = \frac{\ln(1/2)}{8}$$

$$r \approx \frac{-0.6931}{8} \approx -0.0866$$

So the continuous decay rate is about \(-0.0866\), or \(-8.66\%\) per year.

Step 2: Now find the amount after 20 years.

$$A = 500e^{-0.0866(20)}$$

$$A = 500e^{-1.732}$$

Using a calculator, \(e^{-1.732} \approx 0.1768\).

$$A \approx 500(0.1768)$$

$$A \approx 88.4$$

Answer: About 88.4 grams remain after 20 years.

This example shows that half-life problems can still be solved with the same formula. Sometimes you are given the rate directly, and sometimes you must find it first.

Solving for time is another important skill. Sometimes you know the starting amount and ending amount, and you need to find how long the process takes.

Suppose a substance follows \(A = Pe^{rt}\). To solve for \(t\):

1. Divide both sides by \(P\): \(\frac{A}{P} = e^{rt}\)
2. Take the natural logarithm of both sides: \(\ln\left(\frac{A}{P}\right) = rt\)
3. Solve for \(t\):

$$t = \frac{\ln(A/P)}{r}$$

This is especially useful in questions about how long it takes for an investment to reach a certain value or how long it takes for a substance to decay to a certain amount.

Quick example of solving for time

An investment of \(\$1000\) grows continuously at 6% per year. How long will it take to reach \(\$1500\)?

Use

$$1500 = 1000e^{0.06t}$$

Divide by 1000:

$$1.5 = e^{0.06t}$$

Take the natural logarithm:

$$\ln(1.5) = 0.06t$$

$$t = \frac{\ln(1.5)}{0.06}$$

$$t \approx \frac{0.4055}{0.06} \approx 6.76$$

Answer: It will take about 6.76 years.

Common mistakes to avoid

• Forgetting to change percent to decimal: 7% must be written as \(0.07\), not 7.
• Using a positive rate for decay: decay rates must be negative in \(A = Pe^{rt}\).
• Mixing time units: if \(r\) is per year, then \(t\) must be in years.
• Entering calculator expressions incorrectly: make sure the whole exponent is inside, such as \(e^{0.05 \cdot 3}\).
• Rounding too early: keep several decimal places until the final answer.

How to tell when to use this model

• If the problem says compounded continuously, use \(A = Pe^{rt}\).
• If the problem says grows continuously or decays continuously, use \(A = Pe^{rt}\).
• If the problem gives a half-life, you can still use this model after finding or using the decay rate.

Summary

Continuous growth and decay are modeled by the formula $$A = Pe^{rt}$$ where \(P\) is the initial amount, \(r\) is the rate as a decimal, and \(t\) is time. Positive \(r\) means growth, and negative \(r\) means decay.

These models are useful for interest, population change, and radioactive decay. To solve problems correctly, pay attention to the sign of the rate, match time units, and use logarithms when you need to solve for time or rate.