Logarithmic Functions and Equations

Lesson: The Logarithmic Inverse

In earlier work with exponents, you learned how to evaluate expressions like \(2^3 = 8\) or \(10^2 = 100\). But sometimes in maths, we know the result and want to find the exponent. For example, if \(2^x = 8\), what is \(x\)? Since \(2^3 = 8\), the answer is \(x = 3\).

This idea leads to logarithms. A logarithm tells us the exponent needed to produce a certain value. In this way, logarithms are the inverse of exponential functions.

In this lesson, you will learn what it means for logarithms to be inverses of exponentials, how to switch between exponential and logarithmic form, and how to interpret logarithms correctly.

1. What does “inverse” mean?

Two operations are inverses if one undoes the other. For example:

• Addition and subtraction are inverses.
• Multiplication and division are inverses.
• Squaring and square roots are inverses.

In the same way, exponentials and logarithms are inverses.

If

$$b^y = x$$

then the equivalent logarithmic statement is

$$\log_b(x) = y$$

This is the key definition of a logarithm.

Read it like this: “log base \(b\) of \(x\) equals \(y\).”

It means: the exponent on base \(b\) that gives \(x\) is \(y\).

2. The definition of a logarithm

The formal definition is:

$$\log_b(x)=y \quad \text{if and only if} \quad b^y=x$$

This means you can move back and forth between the two forms whenever needed.

To use logarithms correctly, the base must satisfy:

• \(b > 0\)
• \(b \ne 1\)

Also, the input of a logarithm must be positive:

• \(x > 0\)

So expressions like \(\log_2(8)\) are valid, but \(\log_2(-8)\) is not.

3. Converting between exponential and logarithmic form

The most important skill in this topic is converting between the two forms.

Exponential form:

$$b^y=x$$

Logarithmic form:

$$\log_b(x)=y$$

Notice how each part matches:

• The base stays the same.
• The exponent becomes the answer to the logarithm.
• The result of the exponential becomes the input of the logarithm.

A helpful way to remember this is:

base^answer = number

So

$$\log_b(x)=y \iff b^y=x$$

4. Understanding what a logarithm asks

When you see a logarithm such as \(\log_3(81)\), do not think of it as multiplication or division. Instead ask:

“3 to what power equals 81?”

Since

$$3^4=81$$

we have

$$\log_3(81)=4$$

That is the main meaning of a logarithm.

5. Worked Example 1: From exponential form to logarithmic form

Write \(2^5 = 32\) in logarithmic form.

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

• Base: \(2\)
• Exponent: \(5\)
• Result: \(32\)

Step 2: Use the pattern \(b^y=x \iff \log_b(x)=y\).

So

$$\log_2(32)=5$$

Answer: \(\log_2(32)=5\)

6. Worked Example 2: From logarithmic form to exponential form

Write \(\log_4(64)=3\) in exponential form.

Step 1: Identify the base, input, and output.

• Base: \(4\)
• Input: \(64\)
• Output: \(3\)

Step 2: Rewrite using \(b^y=x\).

So

$$4^3=64$$

Answer: \(4^3=64\)

7. Worked Example 3: Evaluating a logarithm

Find \(\log_5(125)\).

This asks:

“5 to what power equals 125?”

Since

$$5^3=125$$

it follows that

$$\log_5(125)=3$$

Answer: \(3\)

8. Worked Example 4: Solving a simple exponential equation using logarithmic thinking

Solve \(3^x = 27\).

Method 1: Use exponent knowledge

Since

$$3^3=27$$

we get

$$x=3$$

Method 2: Rewrite using logarithms

Because \(3^x=27\), we can write

$$\log_3(27)=x$$

Now ask: “3 to what power gives 27?” Since \(3^3=27\),

$$x=3$$

This shows how logarithms help describe unknown exponents.

9. Special logarithm values

Some logarithms appear often and are useful to remember.

• \(\log_b(1)=0\) because \(b^0=1\)
• \(\log_b(b)=1\) because \(b^1=b\)

Examples:

• \(\log_7(1)=0\)
• \(\log_9(9)=1\)
• \(\log_2(1)=0\)

10. How inverse relationships work

Since logarithms and exponentials undo each other, the following relationships are true:

$$\log_b(b^x)=x$$ $$b^{\log_b(x)}=x$$

These work because one operation reverses the other.

For example:

$$\log_2(2^6)=6$$ $$10^{\log_{10}(1000)}=1000$$

You do not need to think of these as complicated rules. They simply show that inverse operations cancel each other.

11. Common mistakes to avoid

• Mixing up the base and the result.
  For \(\log_2(8)=3\), the base is \(2\), not \(8\).
• Reading a logarithm incorrectly.
  \(\log_3(81)\) means “3 to what power gives 81?”
• Using negative or zero inputs.
  \(\log_b(x)\) is only defined when \(x>0\).
• Forgetting that logarithms give exponents.
  The answer to a logarithm is a power.

12. Quick practice ideas

Try asking yourself these questions:

• \(\log_2(16)= ?\)
• \(\log_{10}(1000)= ?\)
• Write \(5^4=625\) in logarithmic form.
• Write \(\log_6(36)=2\) in exponential form.

Answers:

• \(\log_2(16)=4\)
• \(\log_{10}(1000)=3\)
• \(\log_5(625)=4\)
• \(6^2=36\)

13. Summary

A logarithm is the inverse of an exponential. The statement \(\log_b(x)=y\) means exactly the same thing as \(b^y=x\).

When evaluating a logarithm, always ask: “The base raised to what power equals this number?” If you can convert smoothly between logarithmic and exponential form, you have understood the key idea of the logarithmic inverse.

Common and Natural Logarithms are two very important types of logarithms you will use often in Algebra. They help us rewrite exponential expressions, solve equations with variables in exponents, and model real-world situations such as growth and decay.

Before learning the special types, remember the main idea of a logarithm: a logarithm answers the question, “What power do I raise the base to in order to get this number?”

In general,

$$ \log_b a = c \quad \text{means} \quad b^c = a $$

This shows that logarithms and exponential functions are inverses of each other.

For example,

$$ \log_2 8 = 3 $$

because

$$ 2^3 = 8 $$

In this lesson, we focus on the two most common logarithms used in 11th Grade Maths:

• Common logarithm: base 10
• Natural logarithm: base \(e\)

1. Common Logarithms

A common logarithm is a logarithm with base 10.

$$ \log x = \log_{10} x $$

Notice that when no base is written, in most school mathematics this means base 10.

For example,

$$ \log 100 = 2 $$

because

$$ 10^2 = 100 $$

Another example is

$$ \log 1000 = 3 $$

because

$$ 10^3 = 1000 $$

Common logs are especially helpful when working with powers of 10 and when using a calculator.

2. Natural Logarithms

A natural logarithm is a logarithm with base \(e\).

$$ \ln x = \log_e x $$

The number \(e\) is an irrational number approximately equal to

$$ e \approx 2.718 $$

You do not need to memorize many decimal places. Just know that \(e\) is a special constant, similar to how \(\pi\) is special in geometry.

For example,

$$ \ln e = 1 $$

because

$$ e^1 = e $$

Also,

$$ \ln 1 = 0 $$

because

$$ e^0 = 1 $$

3. Key Facts to Know

• \(\log x\) means base 10.
• \(\ln x\) means base \(e\).
• Logarithms are only defined when the input is positive.

That means:

$$ \log x \text{ is defined only if } x>0 $$

and

$$ \ln x \text{ is defined only if } x>0 $$

So expressions like \(\log(-5)\) and \(\ln(0)\) are not defined in real numbers.

4. Important Exact Values

Some logarithms can be found exactly without a calculator.

For common logarithms:

• \(\log 1 = 0\), because \(10^0=1\)
• \(\log 10 = 1\), because \(10^1=10\)
• \(\log 100 = 2\), because \(10^2=100\)
• \(\log 1000 = 3\), because \(10^3=1000\)

For natural logarithms:

• \(\ln 1 = 0\), because \(e^0=1\)
• \(\ln e = 1\), because \(e^1=e\)
• \(\ln e^2 = 2\), because \(e^2=e^2\)
• More generally, \(\ln(e^k)=k\)

5. Using a Calculator

Most calculators have a log button and an ln button.

• Use log for common logarithms.
• Use ln for natural logarithms.

For example, a calculator gives approximate values such as

$$ \log 7 \approx 0.8451 $$

because

$$ 10^{0.8451} \approx 7 $$

And

$$ \ln 7 \approx 1.9459 $$

because

$$ e^{1.9459} \approx 7 $$

These are approximate values, not exact values.

6. Exact Value vs Approximate Value

It is important to know when an answer should stay exact and when a decimal approximation is needed.

• If the logarithm can be rewritten using a known power, give an exact value.
• If it cannot be simplified easily, use a calculator approximation.

For example:

• \(\log 100 = 2\) is exact.
• \(\ln e^3 = 3\) is exact.
• \(\log 6 \approx 0.7782\) is approximate.
• \(\ln 6 \approx 1.7918\) is approximate.

7. Logarithms Undo Exponentials

Since logarithms and exponentials are inverses, they cancel each other when the base matches.

For base 10:

$$ \log(10^x)=x $$

and

$$ 10^{\log x}=x \quad \text{for } x>0 $$

For base \(e\):

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

and

$$ e^{\ln x}=x \quad \text{for } x>0 $$

These relationships are very useful for solving equations.

Worked Example 1: Evaluating a Common Logarithm Exactly

Find \(\log 10000\).

Step 1: Rewrite the number as a power of 10.

$$ 10000 = 10^4 $$

Step 2: Use the definition of logarithm.

$$ \log 10000 = \log(10^4)=4 $$

Answer: \(\boxed{4}\)

Worked Example 2: Evaluating a Natural Logarithm Exactly

Find \(\ln\left(\frac{1}{e^2}\right)\).

Step 1: Rewrite the expression using a power of \(e\).

$$ \frac{1}{e^2}=e^{-2} $$

Step 2: Apply the natural log.

$$ \ln(e^{-2})=-2 $$

Answer: \(\boxed{-2}\)

Worked Example 3: Using a Calculator

Find \(\log 35\) to 4 decimal places.

This is not an obvious power of 10, so we use a calculator.

$$ \log 35 \approx 1.5441 $$

Answer: \(\boxed{1.5441}\)

You can check the meaning of this answer:

$$ 10^{1.5441} \approx 35 $$

Worked Example 4: Solving an Exponential Equation

Solve \(e^x = 12\).

The variable is in the exponent, so use the natural logarithm.

Step 1: Take \(\ln\) of both sides.

$$ \ln(e^x)=\ln 12 $$

Step 2: Simplify the left side.

$$ x=\ln 12 $$

Step 3: Approximate if needed.

$$ x \approx 2.4849 $$

Answer: \(\boxed{x=\ln 12 \approx 2.4849}\)

8. Common Mistakes to Avoid

• Forgetting the base: \(\log x\) and \(\ln x\) are not the same.
• Using logs on negative numbers or zero: logarithms only accept positive inputs.
• Confusing exact and approximate answers: \(\ln e = 1\) is exact, but \(\ln 2\) is approximate.
• Thinking log and ln are ordinary variables: they are functions, like square root or absolute value.

9. Quick Comparison: Common Log vs Natural Log

• Common log: \(\log x = \log_{10}x\)
• Natural log: \(\ln x = \log_e x\)
• Calculator buttons: use log for base 10 and ln for base \(e\)
• Exact values: easiest when the number is a power of the base

10. Summary

Common and natural logarithms are special logarithms with bases 10 and \(e\). A common logarithm is written as \(\log x\), and a natural logarithm is written as \(\ln x\).

They are inverses of exponential functions, which means they help us undo exponents. You should be able to recognize exact values such as \(\log 100=2\) and \(\ln e^3=3\), and use a calculator to approximate values like \(\log 7\) or \(\ln 12\).

Most importantly, remember that logarithms are only defined for positive numbers, and always pay attention to whether the question asks for an exact value or a decimal approximation.

Properties of Logarithms

Logarithms are closely connected to exponents. In fact, a logarithm answers the question: "What exponent gives this result?"

For example, in the expression \(\log_2 8 = 3\), the logarithm tells us that the exponent on 2 must be 3 because \(2^3 = 8\).

Because logarithms are the inverse of exponential functions, they follow patterns that help us rewrite and simplify expressions. These patterns are called the properties of logarithms.

In this lesson, you will learn how to use the three main properties:

• Product Property
• Quotient Property
• Power Property

You will also learn how to expand a logarithmic expression and how to condense several logarithms into one.

Before using logarithms, remember the basic restrictions:

• The base must be positive and not equal to 1.
• The number inside the logarithm, called the argument, must be positive.

So expressions like \(\log_3(-5)\) are not defined in the real number system.

1. Product Property of Logarithms

The product property says that the logarithm of a product can be written as the sum of two logarithms.

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

This property works when \(M>0\), \(N>0\), and the base \(b\) is valid.

This means that multiplying inside a logarithm becomes addition outside the logarithm.

Example:

$$\log_5(2x)=\log_5 2+\log_5 x$$

This is helpful when you want to break a log expression into simpler parts.

2. Quotient Property of Logarithms

The quotient property says that the logarithm of a quotient can be written as the difference of two logarithms.

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

Again, this only works if \(M>0\) and \(N>0\).

This means that division inside a logarithm becomes subtraction outside the logarithm.

Example:

$$\log_4\left(\frac{x}{7}\right)=\log_4 x-\log_4 7$$

3. Power Property of Logarithms

The power property says that an exponent inside a logarithm can be brought in front as a coefficient.

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

This is often the most useful property when expanding logarithms.

Example:

$$\log_3(x^4)=4\log_3 x$$

Notice that the exponent becomes a multiplier in front of the logarithm.

Important note: The power property works in one direction and the other direction. You can move an exponent out front, and you can also move a coefficient back in as an exponent when condensing.

Common Mistake to Avoid

Do not think that logarithms distribute over addition or subtraction.

These are not true:

• \(\log_b(M+N) \ne \log_b M+\log_b N\)
• \(\log_b(M-N) \ne \log_b M-\log_b N\)

The product property only works for multiplication, and the quotient property only works for division.

For example,

$$\log(2+3) = \log 5$$

but

$$\log 2 + \log 3 = \log 6$$

Since \(\log 5 \ne \log 6\), the expression cannot be split across addition.

Expanding Logarithmic Expressions

To expand means to rewrite one logarithm as a sum or difference of simpler logarithms.

When expanding, follow this general order:

1. Use the product or quotient property to split the expression.
2. Use the power property to move exponents out front.
3. Simplify if possible.

Worked Example 1: Expand a simple product

Expand \(\log_2(7x)\).

Step 1: Notice that \(7x\) is a product.

Use the product property:

$$\log_2(7x)=\log_2 7+\log_2 x$$

Final answer:

$$\boxed{\log_2 7+\log_2 x}$$

Worked Example 2: Expand an expression with a quotient and a power

Expand \(\log_3\left(\frac{x^2}{5y}\right)\).

Step 1: The main operation is division, so use the quotient property.

$$\log_3\left(\frac{x^2}{5y}\right)=\log_3(x^2)-\log_3(5y)$$

Step 2: Expand each part.

For \(\log_3(x^2)\), use the power property:

$$\log_3(x^2)=2\log_3 x$$

For \(\log_3(5y)\), use the product property:

$$\log_3(5y)=\log_3 5+\log_3 y$$

Step 3: Put everything together.

$$\log_3\left(\frac{x^2}{5y}\right)=2\log_3 x-(\log_3 5+\log_3 y)$$

Now remove the parentheses carefully:

$$\log_3\left(\frac{x^2}{5y}\right)=2\log_3 x-\log_3 5-\log_3 y$$

Final answer:

$$\boxed{2\log_3 x-\log_3 5-\log_3 y}$$

Condensing Logarithmic Expressions

To condense means to combine several logarithms into a single logarithm.

When condensing, do the reverse of expanding:

1. Use the power property backward to move coefficients inside as exponents.
2. Use addition to combine with the product property.
3. Use subtraction to combine with the quotient property.

Worked Example 3: Condense a sum of logarithms

Condense \(\log_4 x + \log_4 y\).

Since the logs have the same base and are being added, use the product property backward:

$$\log_4 x + \log_4 y = \log_4(xy)$$

Final answer:

$$\boxed{\log_4(xy)}$$

Worked Example 4: Condense a more complex expression

Condense \(3\log_2 x + \log_2 y - 2\log_2 z\).

Step 1: Move the coefficients inside as exponents.

$$3\log_2 x = \log_2(x^3)$$ $$2\log_2 z = \log_2(z^2)$$

So the expression becomes:

$$\log_2(x^3)+\log_2 y-\log_2(z^2)$$

Step 2: Combine the added logs using the product property.

$$\log_2(x^3)+\log_2 y=\log_2(x^3y)$$

Step 3: Use the quotient property for the subtraction.

$$\log_2(x^3y)-\log_2(z^2)=\log_2\left(\frac{x^3y}{z^2}\right)$$

Final answer:

$$\boxed{\log_2\left(\frac{x^3y}{z^2}\right)}$$

How to Know Which Property to Use

• If you see multiplication inside one logarithm, use the product property.
• If you see division inside one logarithm, use the quotient property.
• If you see an exponent inside one logarithm, use the power property.
• If you are combining several logarithms into one, work backward with these same properties.

Important Conditions

Whenever you expand or condense, the arguments of the logarithms must stay positive.

For example, if you write \(\log(xy)\), then both the logarithm expression and its parts must make sense in the real number system. In many 11th Grade problems, it is usually understood that the variables represent positive values unless stated otherwise.

Quick Check

Decide whether each statement is true or false:

• \(\log_b(MN)=\log_b M+\log_b N\) → True
• \(\log_b\left(\frac{M}{N}\right)=\log_b M-\log_b N\) → True
• \(\log_b(M^p)=p\log_b M\) → True
• \(\log_b(M+N)=\log_b M+\log_b N\) → False

Summary

The properties of logarithms help you rewrite expressions in simpler forms. The product property changes multiplication into addition, the quotient property changes division into subtraction, and the power property moves exponents out front.

When you expand, you break one logarithm into several smaller ones. When you condense, you combine several logarithms into one. Just remember: logarithms do not split across addition or subtraction inside the argument.

Change of Base Formula is a useful logarithm rule that helps us evaluate logarithms when the base is not one of the buttons on a calculator.

Most calculators have buttons for common logarithms, written as \(\log x\), which means base 10, and natural logarithms, written as \(\ln x\), which means base \(e\).

But what if you need to find something like \(\log_2 7\) or \(\log_5 18\)? Since calculators usually do not have a button for base 2 or base 5 logs, we use the change of base formula.

This lesson will show you what the formula is, why it works, how to use it, and how it helps solve logarithmic problems.

First, remember what a logarithm means.

If

$$\log_b a = c$$

then this means

$$b^c = a$$

In words, a logarithm tells you the exponent you put on the base \(b\) to get \(a\).

For example,

$$\log_2 8 = 3$$

because

$$2^3 = 8$$

The base of a logarithm must be positive and not equal to 1, and the number inside the logarithm must be positive.

The change of base formula says:

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

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 may use whichever is easier on your calculator.

Why does this work?

Suppose

$$x = \log_b a$$

By the definition of logarithms, this means

$$b^x = a$$

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

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

Use the power rule for logarithms:

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

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 same reasoning works with \(\ln\), so

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

Important idea: change of base lets you rewrite a logarithm in any base using a base your calculator knows.

For example, instead of trying to directly calculate \(\log_3 20\), you rewrite it as

$$\log_3 20 = \frac{\log 20}{\log 3}$$

or

$$\log_3 20 = \frac{\ln 20}{\ln 3}$$

When should you use the change of base formula?

• When the logarithm has a base other than 10 or \(e\)
• When you need a decimal approximation
• When solving equations involving unusual logarithm bases

Worked Example 1: Evaluate a logarithm with an unusual base

Find \(\log_2 7\) to three decimal places.

Use the change of base formula:

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

Now use a calculator:

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

So,

$$\log_2 7 \approx 2.807$$

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

Worked Example 2: Use natural logs instead

Find \(\log_5 18\) to three decimal places.

Use the \(\ln\) version:

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

Using a calculator,

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

So,

$$\log_5 18 \approx 1.796$$

This is reasonable because \(5^1 = 5\) and \(5^2 = 25\), so the answer should be between 1 and 2.

Worked Example 3: Solve an exponential equation

Solve

$$3^x = 11$$

We want the exponent \(x\), so rewrite in logarithmic form:

$$x = \log_3 11$$

Now apply change of base:

$$x = \frac{\log 11}{\log 3}$$

Using a calculator,

$$x \approx \frac{1.0414}{0.4771} \approx 2.183$$

So the solution is

$$x \approx 2.183$$

You can check that this makes sense because \(3^2 = 9\) and \(3^3 = 27\), so the exponent needed to get 11 should be a little more than 2.

Worked Example 4: Solve a logarithmic equation

Solve

$$\log_4 x = 3.5$$

Rewrite in exponential form:

$$x = 4^{3.5}$$

You could evaluate this directly, or rewrite \(3.5\) as \(\frac{7}{2}\):

$$x = 4^{7/2}$$

Since \(4^{1/2} = 2\),

$$4^{7/2} = (4^{1/2})^7 = 2^7 = 128$$

So,

$$x = 128$$

Now let us connect this to change of base. If you instead wanted to check the value using logs, you could write

$$3.5 = \log_4 x = \frac{\log x}{\log 4}$$

This shows how change of base can help when working with logarithmic equations.

Common mistakes to avoid

• Reversing the formula: \(\log_b a = \frac{\log a}{\log b}\), not \(\frac{\log b}{\log a}\).
• Mixing log and ln incorrectly: You may use all \(\log\) or all \(\ln\), but do not mix them in the same fraction.
• Forgetting calculator parentheses: Type the numerator and denominator carefully, for example \((\log 7)/(\log 2)\).
• Ignoring reasonableness: Estimate between which powers the number lies to see if your answer makes sense.

Helpful pattern to remember

If you see a logarithm with a base your calculator does not have, use:

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

Think of it as:

log of the number divided by log of the base.

Quick practice ideas

1. Find \(\log_6 20\) using common log.
2. Find \(\log_7 3\) using natural log.
3. Solve \(2^x = 15\).
4. Solve \(\log_5 x = 2.4\).

Brief Summary

The change of base formula lets you rewrite logarithms in a form that is easier to calculate:

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

This is especially helpful for evaluating logs with unusual bases and for solving exponential equations. Always make sure the argument of the logarithm is positive, and check whether your answer is reasonable by comparing nearby powers.

Solving Logarithmic Equations

Logarithmic equations are equations in which the variable appears inside a logarithm. To solve them, we use the fact that a logarithm is the inverse of an exponential expression.

For example, if $$\log_b(x)=y,$$ then this means $$b^y=x.$$ This relationship is the key idea behind solving logarithmic equations.

In this lesson, you will learn how to:

• rewrite logarithmic equations in exponential form,
• use logarithm properties to combine or simplify expressions,
• solve for the variable, and
• check for extraneous solutions, which are answers that do not actually work in the original equation.

1. Important Logarithm Facts

Before solving equations, remember these basic facts:

• A logarithm asks: To what power must the base be raised?
• For $$\log_b(x)$$ to exist, the input must be positive: $$x>0.$$
• The base must be positive and not equal to 1.

This means that when solving logarithmic equations, you must always check that every log input is positive.

2. Common Logarithm Properties

These properties are very useful when solving equations:

• 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 log expressions have the same base, and the inputs must be positive.

3. Main Strategy for Solving Logarithmic Equations

There are two common types of logarithmic equations.

Type A: A single logarithm equals a number

Example form: $$\log_b(f(x))=c$$

To solve:

1. Rewrite in exponential form: $$f(x)=b^c$$
2. Solve the resulting equation.
3. Check that the solution makes the log input positive.

Type B: Logs on one or both sides

Example form: $$\log_b(A)+\log_b(B)=\log_b(C)$$

To solve:

1. Use log properties to combine logarithms if needed.
2. If possible, get a single log on each side.
3. Use the one-to-one property of logarithms: if $$\log_b(M)=\log_b(N),$$ then $$M=N.$$
4. Solve the resulting equation.
5. Check all solutions in the original equation.

4. Why Checking Solutions Matters

When solving logarithmic equations, algebra steps can sometimes produce answers that make a log input zero or negative. But logarithms are only defined for positive inputs.

That is why you must check every answer in the original equation, not just in a simplified version.

Worked Example 1: Rewriting in Exponential Form

Solve:

$$\log_3(x)=2$$

Rewrite in exponential form:

$$x=3^2$$

So,

$$x=9$$

Check the domain: the input of the log is $$x$$, and $$9>0$$, so the solution is valid.

Answer: $$x=9$$

Worked Example 2: Solve a Log Equation with an Expression Inside

Solve:

$$\log_5(2x-1)=3$$

Rewrite in exponential form:

$$2x-1=5^3$$

$$2x-1=125$$

$$2x=126$$

$$x=63$$

Now check the original log input:

$$2(63)-1=125$$

Since $$125>0$$, the solution is valid.

Answer: $$x=63$$

Worked Example 3: Combining Logarithms

Solve:

$$\log_2(x)+\log_2(x-6)=3$$

First, note the domain restrictions:

• $$x>0$$
• $$x-6>0 \Rightarrow x>6$$

Now combine the logarithms using the product rule:

$$\log_2(x(x-6))=3$$

Rewrite in exponential form:

$$x(x-6)=2^3$$

$$x^2-6x=8$$

$$x^2-6x-8=0$$

Solve the quadratic equation:

$$x=\frac{6\pm\sqrt{36+32}}{2}$$

$$x=\frac{6\pm\sqrt{68}}{2}$$

$$x=3\pm\sqrt{17}$$

Now check against the domain $$x>6$$:

• $$3+\sqrt{17}\approx 7.12$$, which works
• $$3-\sqrt{17}\approx -1.12$$, which does not work

Answer: $$x=3+\sqrt{17}$$

Worked Example 4: Logs on Both Sides and an Extraneous Solution

Solve:

$$\log(x+1)=\log(3x-7)$$

Because the logs have the same base, if they are equal, then their inputs must be equal:

$$x+1=3x-7$$

Solve:

$$1=2x-7$$

$$8=2x$$

$$x=4$$

Now check in the original equation:

• $$x+1=5>0$$
• $$3x-7=12-7=5>0$$

So the solution is valid.

Answer: $$x=4$$

Now look at a quick example where a solution must be rejected.

Solve:

$$\log(x-2)=\log(5-x)$$

Set the arguments equal:

$$x-2=5-x$$

$$2x=7$$

$$x=\frac{7}{2}$$

Check the domain:

• $$x-2=\frac{3}{2}>0$$
• $$5-x=\frac{3}{2}>0$$

This one actually works. So let us see a case that fails.

Solve:

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

Set arguments equal:

$$x-4=2-x$$

$$2x=6$$

$$x=3$$

Now check the original equation:

• $$x-4=3-4=-1$$, not allowed
• $$2-x=2-3=-1$$, not allowed

So $$x=3$$ is an extraneous solution.

Answer: no solution

5. A Step-by-Step Checklist

When solving a logarithmic equation, use this checklist:

1. Look at the inputs of all logarithms and note any domain restrictions.
2. Use logarithm properties to combine or simplify if needed.
3. Rewrite in exponential form, or set the arguments equal if the logs have the same base on both sides.
4. Solve the resulting algebraic equation.
5. Check every solution in the original equation.
6. Reject any solution that makes a log input zero or negative.

6. Common Mistakes to Avoid

• Forgetting the domain: log inputs must always be positive.
• Using log rules incorrectly: for example, $$\log(a+b) \neq \log a + \log b$$
• Not checking answers: even correct algebra can produce invalid log solutions.
• Mixing bases: log properties and equality rules depend on the logs having the same base.

7. Final Summary

To solve logarithmic equations, first simplify the logs using log properties if necessary. Then rewrite the equation in exponential form or, when two logs of the same base are equal, set their inputs equal.

After solving, always check your answer in the original equation. This is important because logarithms only accept positive inputs, so some algebraic answers may be extraneous and must be rejected.

Solving Exponential Equations using Logarithms

In many equations, the variable appears in a normal position, such as in \(2x+3=11\). But in an exponential equation, the variable is in the exponent, like \(2^x=7\). These equations cannot usually be solved with basic algebra alone.

This is where logarithms become useful. Since logarithms are the inverse of exponential functions, they allow us to “bring the exponent down” so we can solve for the variable.

In this lesson, you will learn how to solve exponential equations by taking the logarithm of both sides, using either the common logarithm \((\log)\) or the natural logarithm \((\ln)\).

1. Why logarithms help

Remember that exponential and logarithmic forms are inverses of each other.

For example,

$$2^3=8$$

can be written in logarithmic form as

$$\log_2 8=3$$

This means that if a variable is trapped in the exponent, a logarithm can help uncover it.

2. The basic idea

Suppose we want to solve

$$3^x=10$$

There is no simple way to rewrite 10 as a power of 3. So we take the logarithm of both sides:

$$\log(3^x)=\log(10)$$

Now use the power property of logarithms:

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

So,

$$x\log(3)=\log(10)$$

Now solve for \(x\):

$$x=\frac{\log(10)}{\log(3)}$$

Since \(\log(10)=1\),

$$x=\frac{1}{\log(3)}\approx 2.096$$

You would get the same answer using natural logs:

$$\ln(3^x)=\ln(10)$$ $$x\ln(3)=\ln(10)$$ $$x=\frac{\ln(10)}{\ln(3)}\approx 2.096$$

Important: You can use either \(\log\) or \(\ln\), as long as you use the same one on both sides.

3. Steps for solving exponential equations using logarithms

1. Isolate the exponential expression if needed.

2. Take \(\log\) or \(\ln\) of both sides.

3. Use the power property to move the exponent in front.

4. Solve the resulting algebraic equation.

5. Use a calculator if needed to find a decimal answer.

4. Worked Examples

Example 1: Solve \(5^x=13\)

This cannot be solved by rewriting 13 as a power of 5, so use logarithms.

$$\log(5^x)=\log(13)$$

Apply the power property:

$$x\log(5)=\log(13)$$

Now divide both sides by \(\log(5)\):

$$x=\frac{\log(13)}{\log(5)}$$

Using a calculator,

$$x\approx 1.593$$

So the solution is

$$x\approx 1.593$$

Example 2: Solve \(2^{3x}=20\)

The variable is in the exponent, so take the log of both sides:

$$\log(2^{3x})=\log(20)$$

Use the power property:

$$3x\log(2)=\log(20)$$

Now solve for \(x\):

$$x=\frac{\log(20)}{3\log(2)}$$

Using a calculator,

$$x\approx 1.441$$

So the solution is

$$x\approx 1.441$$

Example 3: Solve \(7\cdot 3^x=50\)

First isolate the exponential part by dividing both sides by 7:

$$3^x=\frac{50}{7}$$

Now take the logarithm of both sides:

$$\log(3^x)=\log\left(\frac{50}{7}\right)$$

Use the power property:

$$x\log(3)=\log\left(\frac{50}{7}\right)$$

Solve for \(x\):

$$x=\frac{\log(50/7)}{\log(3)}$$

Using a calculator,

$$x\approx 1.786$$

So the solution is

$$x\approx 1.786$$

Example 4: Solve \(4^{x-1}=9\)

Take the natural log of both sides:

$$\ln(4^{x-1})=\ln(9)$$

Use the power property:

$$(x-1)\ln(4)=\ln(9)$$

Now divide by \(\ln(4)\):

$$x-1=\frac{\ln(9)}{\ln(4)}$$

Add 1 to both sides:

$$x=1+\frac{\ln(9)}{\ln(4)}$$

Using a calculator,

$$x\approx 2.585$$

So the solution is

$$x\approx 2.585$$

5. When should you use logarithms?

Use logarithms when the variable is in the exponent and the two sides cannot easily be written with the same base.

For example:

• \(2^x=8\) can be solved without logs because \(8=2^3\), so \(x=3\).

• \(2^x=7\) should be solved with logs because 7 is not a power of 2.

6. Common mistakes to avoid

• Forgetting to isolate the exponential expression first.
  If the equation is \(5\cdot 2^x=30\), divide by 5 before taking logs.

• Not using the power property correctly.
  \(\log(2^x)=x\log(2)\), not \(\log(2)^x\).

• Using different types of logs on each side.
  Do not mix \(\log\) on one side and \(\ln\) on the other in the same step.

• Rounding too early.
  Keep calculator values as accurate as possible until the final answer.

7. Quick check: compare two methods

Solve \(10^x=6\).

Using common logs:

$$\log(10^x)=\log(6)$$ $$x\log(10)=\log(6)$$

Since \(\log(10)=1\),

$$x=\log(6)\approx 0.778$$

This example is especially nice because the base is 10, which matches the common logarithm.

8. Final summary

To solve an exponential equation using logarithms, first isolate the exponential expression. Then take the same logarithm of both sides and use the power property to move the exponent in front. After that, solve for the variable just like in a regular algebra equation.

Logarithms are powerful because they let you solve equations where the variable is in the exponent, even when the bases do not match nicely. With practice, the process becomes straightforward: isolate, take logs, apply the power rule, and solve.

Logarithmic Graphs and Asymptotes

In this lesson, you will learn how to graph logarithmic functions, identify their vertical asymptotes, describe their domain and range, and understand how transformations change their graphs.

Logarithmic functions are closely connected to exponential functions. In fact, a logarithm is the inverse of an exponential function. This inverse relationship helps explain the shape of a logarithmic graph.

For example, if

$$y=2^x$$

then its inverse is

$$y=\log_2 x$$

Because inverse functions reflect across the line \(y=x\), the graph of \(y=\log_2 x\) is the reflection of \(y=2^x\) across that line.

1. The Basic Logarithmic Graph

A basic logarithmic function has the form

$$y=\log_b x$$

where \(b>0\) and \(b\ne 1\).

The most important facts about this graph are:

• The domain is \(x>0\).
• The range is all real numbers.
• There is a vertical asymptote at \(x=0\).
• The graph passes through \((1,0)\) because \(\log_b 1=0\).
• The graph passes through \((b,1)\) because \(\log_b b=1\).

An asymptote is a line that a graph gets closer and closer to, but does not touch. For logarithmic graphs, the asymptote is vertical.

Why is \(x=0\) an asymptote? A logarithm is only defined when its input is positive. As \(x\) gets closer to \(0\) from the right, the value of \(\log_b x\) decreases without bound.

For example, in \(y=\log_2 x\):

• If \(x=1\), then \(y=0\)
• If \(x=\frac12\), then \(y=-1\)
• If \(x=\frac14\), then \(y=-2\)

As \(x\to 0^+\), \(y\to -\infty\).

2. Shape of the Graph

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

• If \(b>1\), such as \(2\), \(3\), or \(10\), the graph is increasing.
• If \(0<b<1\), such as \(\frac12\), the graph is decreasing.

So:

• \(y=\log_2 x\) increases from left to right.
• \(y=\log_{1/2} x\) decreases from left to right.

Even though the graph changes direction depending on the base, the domain is still \(x>0\), and the vertical asymptote is still \(x=0\).

3. Domain, Range, and Intercepts

For the basic logarithmic function \(y=\log_b x\):

• Domain: \((0,\infty)\)
• Range: \(( -\infty, \infty )\)
• x-intercept: \((1,0)\)
• y-intercept: none, because \(x=0\) is not in the domain

This is important: logarithmic graphs usually do not have a y-intercept unless the graph has been shifted so that \(x=0\) is allowed in the new domain.

4. Transformations of Logarithmic Graphs

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

A transformed logarithmic function often looks like

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

Each part affects the graph in a specific way:

• \(h\): horizontal shift
• \(k\): vertical shift
• \(a\): vertical stretch, compression, or reflection

Horizontal shift and asymptote

In \(y=\log_b(x-h)\), the graph shifts right by \(h\) units if \(h>0\), and left if \(h<0\).

The vertical asymptote moves with the graph. Since the basic asymptote is \(x=0\), the new asymptote becomes

$$x=h$$

The domain also changes. Because the expression inside the logarithm must be positive, we solve

$$x-h>0$$

which gives

$$x>h$$

So the asymptote and domain come directly from the inside expression.

Vertical shift

In \(y=\log_b(x-h)+k\), the \(+k\) moves the graph up or down. This does not change the vertical asymptote or the domain.

Reflection and vertical stretch

If there is a negative sign in front, such as

$$y=-\log_b x$$

the graph is reflected across the \(x\)-axis.

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

5. How to Find the Asymptote and Domain

For any logarithmic function, the key rule is:

The expression inside the logarithm must be greater than 0.

If the function is

$$y=\log_b(\text{inside expression})$$

then solve

$$\text{inside expression} > 0$$

to find the domain.

To find the vertical asymptote, set the inside expression equal to 0.

$$\text{inside expression} = 0$$

This works because the graph approaches the place where the logarithm stops being defined.

Worked Example 1: Graphing a Basic Logarithmic Function

Graph \(y=\log_2 x\). State the domain, range, and asymptote.

Step 1: Identify the parent function.

This is the basic logarithmic function with base \(2\).

Step 2: Find key features.

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

Step 3: Plot a few points.

• If \(x=1\), then \(y=\log_2 1=0\), so \((1,0)\)
• If \(x=2\), then \(y=\log_2 2=1\), so \((2,1)\)
• If \(x=4\), then \(y=\log_2 4=2\), so \((4,2)\)
• If \(x=\frac12\), then \(y=\log_2 \frac12=-1\), so \((\frac12,-1)\)

Step 4: Describe the graph.

The graph increases slowly from left to right, passes through \((1,0)\), and gets closer and closer to the line \(x=0\) without touching it.

Answer:

• Domain: \((0,\infty)\)
• Range: \(( -\infty, \infty )\)
• Vertical asymptote: \(x=0\)

Worked Example 2: Horizontal Shift

Graph \(y=\log_3(x-4)\). State the domain and asymptote.

Step 1: Look at the inside expression.

The function is shifted right 4 units from \(y=\log_3 x\).

Step 2: Find the asymptote.

Set the inside equal to 0:

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

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

Step 3: Find the domain.

The inside must be positive:

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

So the domain is \((4,\infty)\).

Step 4: Find some points.

Choose values so that \(x-4\) is a power of 3.

• If \(x-4=1\), then \(x=5\), and \(y=\log_3 1=0\), so \((5,0)\)
• If \(x-4=3\), then \(x=7\), and \(y=1\), so \((7,1)\)
• If \(x-4=\frac13\), then \(x=\frac{13}{3}\), and \(y=-1\)

Answer:

• Vertical asymptote: \(x=4\)
• Domain: \((4,\infty)\)

Worked Example 3: Reflection and Vertical Shift

Graph \(y=-\log_2 x+3\). Describe the transformations.

Step 1: Start with the parent function.

The parent is \(y=\log_2 x\).

Step 2: Identify transformations.

• The negative sign reflects the graph across the \(x\)-axis.
• The \(+3\) shifts the graph up 3 units.

Step 3: Determine asymptote and domain.

There is no horizontal shift, so the asymptote remains

$$x=0$$

The domain remains

$$x>0$$

Step 4: Plot points.

Use points from \(y=\log_2 x\), reflect them, then shift up.

• \((1,0)\) becomes \((1,3)\)
• \((2,1)\) becomes \((2,2)\)
• \((4,2)\) becomes \((4,1)\)
• \((\frac12,-1)\) becomes \((\frac12,4)\)

Answer:

• Reflection across the \(x\)-axis
• Shift up 3 units
• Domain: \((0,\infty)\)
• Vertical asymptote: \(x=0\)

Worked Example 4: Finding Domain and Asymptote from a More General Function

Consider

$$y=\log_5(2x-6)$$

Find the domain and vertical asymptote.

Step 1: Find the domain.

The inside must be greater than 0:

$$2x-6>0$$ $$2x>6$$ $$x>3$$

So the domain is \((3,\infty)\).

Step 2: Find the asymptote.

Set the inside equal to 0:

$$2x-6=0$$ $$2x=6$$ $$x=3$$

So the vertical asymptote is \(x=3\).

Step 3: Understand the graph.

This graph behaves like a logarithmic curve starting just to the right of \(x=3\). It never crosses the asymptote.

Answer:

• Domain: \((3,\infty)\)
• Vertical asymptote: \(x=3\)

6. Common Mistakes to Avoid

• Using \(x\ge 0\) for the domain. For logarithms, the input must be strictly positive, so the domain is \(x>0\), not \(x\ge 0\).
• Forgetting to shift the asymptote. In \(y=\log_b(x-2)\), the asymptote is \(x=2\), not \(x=0\).
• Mixing up horizontal and vertical shifts. The change inside the logarithm affects the asymptote and domain. The change outside does not.
• Trying to plug in values that make the inside 0 or negative. Those values are not in the domain.

7. Quick Strategy for Any Log Graph

1. Find the expression inside the logarithm.
2. Set it equal to 0 to get the vertical asymptote.
3. Solve it greater than 0 to get the domain.
4. Identify any reflections or shifts.
5. Plot a few easy points.

Summary

A logarithmic function such as \(y=\log_b x\) has domain \(x>0\), range all real numbers, and a vertical asymptote at \(x=0\). If the function is transformed, the asymptote moves according to the expression inside the logarithm. To find the domain, make the inside of the logarithm greater than 0, and to find the vertical asymptote, set the inside equal to 0.

Once you know the asymptote, domain, and transformations, graphing logarithmic functions becomes much easier. Always begin with the parent graph and then apply the changes step by step.

Logarithmic Scales in Science

In science, some quantities can vary by enormous amounts. For example, one earthquake may release far more energy than another, and one sound may be many times more intense than another. Writing these values on an ordinary linear scale can be awkward because the numbers become very large or very small.

To handle this, scientists often use a logarithmic scale. A logarithmic scale is based on logarithms, so equal steps on the scale represent multiplication by the same factor, not addition by the same amount.

Since logarithms are the inverse of exponentials, logarithmic scales are especially useful when quantities grow or shrink exponentially. In this lesson, you will learn how logarithmic scales work and how to interpret three important examples: pH, decibels, and the Richter scale.

1. What does a logarithmic scale mean?

On a linear scale, moving from 2 to 4 means adding 2, and moving from 20 to 22 also means adding 2. The change is measured by equal differences.

On a logarithmic scale, equal steps mean equal multiplicative changes. For example, if a base-10 logarithmic scale increases by 1, the original quantity becomes 10 times as large.

This comes from the definition of logarithms:

$$y = \log_{10}(x) \iff 10^y = x$$

So if one value has log 3 and another has log 4, then the original quantities are:

$$10^3 = 1000 \quad \text{and} \quad 10^4 = 10000$$

Even though the logarithmic values differ by only 1, the actual quantities differ by a factor of 10.

Key idea: On many scientific logarithmic scales, a difference of 1 unit means the actual quantity changes by a factor of 10.

2. Why scientists use logarithmic scales

• They compress very large ranges of values. Huge numbers become easier to compare.
• They show ratios clearly. Multiplying by 10, 100, or 1000 becomes easier to interpret.
• They match how some natural phenomena behave. Many processes in science involve exponential growth or decay.

3. The pH scale

The pH scale measures how acidic or basic a solution is. It is defined by:

$$\text{pH} = -\log_{10}[H^+]$$

Here, \([H^+]\) represents the concentration of hydrogen ions.

The negative sign is important. Because of it, a lower pH means a higher hydrogen ion concentration, so the solution is more acidic.

For example:

• pH 3 is more acidic than pH 4
• A decrease of 1 pH unit means the hydrogen ion concentration becomes 10 times greater
• A decrease of 2 pH units means the concentration becomes 100 times greater

If we solve the pH formula for \([H^+]\), we get:

$$[H^+] = 10^{-\text{pH}}$$

This lets us move back and forth between pH and hydrogen ion concentration.

Worked Example 1: Finding hydrogen ion concentration from pH

A solution has pH 5. Find \([H^+]\).

Use the formula:

$$[H^+] = 10^{-\text{pH}}$$

Substitute \(\text{pH} = 5\):

$$[H^+] = 10^{-5}$$

So the hydrogen ion concentration is:

$$[H^+] = 1 \times 10^{-5}$$

Answer: The hydrogen ion concentration is \(10^{-5}\).

Worked Example 2: Comparing acidity using pH

Solution A has pH 2 and Solution B has pH 6. How many times greater is the hydrogen ion concentration in Solution A?

First, write each concentration:

$$[H^+]_A = 10^{-2}$$

$$[H^+]_B = 10^{-6}$$

Now compare them by dividing:

$$\frac{[H^+]_A}{[H^+]_B} = \frac{10^{-2}}{10^{-6}} = 10^{4} = 10000$$

Answer: Solution A has 10,000 times the hydrogen ion concentration of Solution B.

Important shortcut for pH: If two solutions differ by \(n\) pH units, their hydrogen ion concentrations differ by a factor of \(10^n\).

4. The decibel scale

Sound intensity is often measured using decibels (dB), which form a logarithmic scale. A common formula is:

$$\beta = 10\log_{10}\left(\frac{I}{I_0}\right)$$

Here:

• \(\beta\) is the sound level in decibels
• \(I\) is the sound intensity
• \(I_0\) is a reference intensity

This formula compares a sound’s intensity to a standard reference intensity. Because of the logarithm, decibel values increase much more slowly than the actual intensity.

A very useful fact comes from the formula:

• If intensity becomes 10 times as large, the decibel level increases by 10 dB
• If intensity becomes 100 times as large, the decibel level increases by 20 dB
• If intensity becomes 1000 times as large, the decibel level increases by 30 dB

This is because:

$$10\log_{10}(10) = 10, \quad 10\log_{10}(100) = 20, \quad 10\log_{10}(1000) = 30$$

Worked Example 3: Comparing sound intensities from decibels

One machine produces a sound level of 80 dB, and another produces 100 dB. How many times more intense is the 100 dB sound?

The difference is:

$$100 - 80 = 20 \text{ dB}$$

A 20 dB increase means the intensity is multiplied by 100.

Answer: The 100 dB sound is 100 times more intense.

We can also show this with logarithms. Let the intensity ratio be \(r\). Then:

$$20 = 10\log_{10}(r)$$

Divide by 10:

$$2 = \log_{10}(r)$$

Rewrite in exponential form:

$$r = 10^2 = 100$$

5. The Richter scale

The Richter scale is used to describe earthquake magnitude. It is also logarithmic. A simplified model is:

$$M = \log_{10}\left(\frac{I}{I_0}\right)$$

Here:

• \(M\) is the earthquake magnitude
• \(I\) is the measured intensity
• \(I_0\) is a reference intensity

Because this is a logarithmic scale:

• An increase of 1 in magnitude means the intensity is 10 times greater
• An increase of 2 means the intensity is 100 times greater
• An increase of 3 means the intensity is 1000 times greater

This explains why a magnitude 7 earthquake is far stronger than a magnitude 5 earthquake. The difference is not just 2 units in a simple sense; it means the intensity is 100 times greater.

Worked Example 4: Comparing earthquake intensities

An earthquake of magnitude 6.8 and an earthquake of magnitude 4.8 are recorded. How many times greater is the intensity of the 6.8 earthquake?

Find the difference in magnitudes:

$$6.8 - 4.8 = 2.0$$

A difference of 2 on a base-10 logarithmic scale means a factor of:

$$10^2 = 100$$

Answer: The magnitude 6.8 earthquake is 100 times as intense as the magnitude 4.8 earthquake.

6. A general rule for logarithmic scales

Many scientific logarithmic scales can be understood using this pattern:

$$\text{Scale value} = \log_{10}(\text{quantity or ratio})$$

or sometimes:

$$\text{Scale value} = k\log_{10}(\text{quantity or ratio})$$

where \(k\) is a constant, such as 10 in the decibel formula.

If two measurements differ by \(d\) units on the logarithmic scale, then the original quantities differ by a factor of:

$$10^d$$

If the formula has a coefficient like 10, you first divide by that coefficient.

For decibels, if the sound level difference is \(D\), then:

$$D = 10\log_{10}(r)$$

so

$$\log_{10}(r) = \frac{D}{10}$$

and therefore:

$$r = 10^{D/10}$$

7. Common mistakes to avoid

• Thinking a difference of 1 means “1 more”. On a logarithmic scale, a difference of 1 usually means a factor of 10.
• Forgetting the negative sign in pH. Lower pH means greater hydrogen ion concentration.
• Treating decibels like a simple linear measure. A 20 dB increase does not mean “20 more intensity”; it means 100 times the intensity.
• Ignoring the coefficient in formulas. In the decibel formula, the 10 in front of the logarithm matters.

8. Quick interpretation guide

• pH: Lower pH = more acidic. A change of 1 pH unit means a factor of 10 in hydrogen ion concentration.
• Decibels: A 10 dB increase means 10 times the sound intensity.
• Richter scale: A 1-unit increase means 10 times the earthquake intensity.

Brief Summary

Logarithmic scales are used in science when quantities vary by very large factors. Instead of measuring equal differences, they measure equal ratios. On these scales, a small change in the scale value can represent a huge change in the actual quantity.

The pH scale, decibel scale, and Richter scale all use logarithms to make large ranges easier to understand. To solve problems on these scales, pay close attention to how many units the scale changes and whether the formula includes a coefficient or a negative sign.