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Logarithmic Expressions

Lesson ~11 min read 8 MCQs

In simple terms: In simple terms, logarithms are a way to find the hidden exponent needed to turn one number into another. For example, the logarithm of 100 (base 10) is 2, because 10 to the power of 2 is 100.

Why this matters

Have you ever heard a news report about an earthquake in California and another in Alaska? The reporter might say the California quake was a magnitude 5.0, while the one in Alaska was a 6.0. You might think, "Okay, so the Alaska one was just a little bit stronger." But in reality, it was ten times stronger. A magnitude 7.0 would be one hundred times stronger than the 5.0.

This is the power of a logarithmic scale. It helps us make sense of numbers that grow incredibly fast. Instead of a ruler where each inch is the same, a logarithmic ruler's marks might go 1, 10, 100, 1000. Understanding logarithms is the key to unlocking these kinds of powerful relationships, which appear everywhere from earthquake intensity to sound levels and even the brightness of stars. Today, we'll learn how to read and work with these essential expressions.

Comparing linear vs. logarithmic scales for earthquake magnitudes.

Concept overview

flowchart TD
    A[Start: Evaluate log_b(c)] --> B{Set it equal to x<br>log_b(c) = x};
    B --> C[Rewrite in exponential form<br>b^x = c];
    C --> D{Solve for x};
    D --> E[Is c a direct power of b?<br>e.g., 2^x = 8];
    E -- Yes --> F[Solve directly<br>x = 3];
    E -- No --> G{Find a common base<br>e.g., 64^x = 4 --> (4^3)^x = 4^1};
    G --> H[Use exponent rules to solve<br>3x = 1 --> x = 1/3];
    F --> I[End: The value is x];
    H --> I;

Core explanation

Welcome to the world of logarithms! It can feel like learning a new language at first, but I promise it's more straightforward than it looks.

The Core Idea: Logs are Exponents in Disguise

Think of a logarithm as a way to solve for a missing exponent.

You're already comfortable with an expression like 2^3 = 8. Here, you have the base (2), the exponent (3), and the result (8).

A logarithm just rearranges this relationship to help you find the exponent. We would write it like this:

log₂(8) = 3

Let's break that down:

  • log: This is just the name of the function.
  • ₂: This little subscript number is the base. It's the same base as in the exponential version (2^3).
  • (8): This is the argument. It's the result we are trying to get to.
  • = 3: This is the value of the logarithm, and here's the most important part: the value of a log is always the exponent.

So, when you see log₂(8), the question you should ask yourself is: "What power do I need to raise the base (2) to, in order to get the argument (8)?" The answer is 3.

This gives us our fundamental rule:

log_b(c) = a is the exact same thing as b^a = c

The fundamental equivalence between logarithmic and exponential forms.

This "rewriting" trick is your single most powerful tool for this topic. Let's see it in action.

Evaluating a Basic Logarithm

Let's evaluate log₂(8).

  1. 1
    Set it equal to a variable
    Let's call our unknown answer 'a'. log₂(8) = a
  2. 2
    Rewrite it in exponential form
    The base 2 swings over to become the base of the exponent a, and the argument 8 is the result. 2^a = 8
  3. 3
    Solve for the variable
    What power of 2 gives you 8? You know that 2 × 2 × 2 = 8, or 2³ = 8. So, a = 3.

Therefore, log₂(8) = 3.

The "Common" Logarithm: What if there's no base?

Sometimes you'll see a logarithm written without a base, like this:

log(100)

So, log(100) is just shorthand for log₁₀(100).

Why base 10? Because our entire number system is base-10. It's the "common" way we count.

Let's evaluate it:

  1. 1
    Problem
    log(100)
  2. 2
    Rewrite with the implied base
    log₁₀(100) = a
  3. 3
    Rewrite in exponential form
    10^a = 100
  4. 4
    Solve
    What power of 10 gives you 100? 10² = 100. So, a = 2.

Therefore, log(100) = 2.

When the Answer Isn't a Whole Number

Logarithms can also have negative or fractional answers. Don't let it throw you. The process is the same.

Let's evaluate log₃(1/9).

  1. 1
    Set it equal to a variable
    log₃(1/9) = a
  2. 2
    Rewrite in exponential form
    3^a = 1/9
  3. 3
    Solve
    This looks tricky. But remember your exponent rules. How can you get a fraction? With a negative exponent! We know that 3² = 9. Therefore, 3⁻² = 1/3² = 1/9. So, a = -2.

Therefore, log₃(1/9) = -2.

Exact Values vs. Calculator Estimates

We've been able to solve all these examples using basic arithmetic because the arguments were perfect powers of the base. We can find the exact value for log₂(8).

But what about log₂(9)? If we rewrite it as 2^a = 9, we get stuck. We know 2³ = 8 and 2⁴ = 16. So the answer must be somewhere between 3 and 4, but it's not a nice, clean number.

This is a case where the value isn't "readily accessible." For an expression like this on the AP exam, you would use a calculator to find an estimated value (which is approximately 3.17).

Back to Logarithmic Scales

Let's tie this back to our earthquake example. The Richter scale is a base-10 logarithmic scale. The magnitude is the value of the logarithm.

  • Magnitude 5
    log(x) = 5, which means the energy x is 10⁵.
  • Magnitude 6
    log(y) = 6, which means the energy y is 10⁶.

Worked examples

Let's walk through a few examples together. The key in every single one is to rewrite the expression.

Example 1

Evaluating a Logarithm with an Integer Solution

Problem: Evaluate log₅(125).

Step 1: Set the expression equal to a variable. Let's call our unknown answer x. This gives us a full equation to work with. log₅(125) = x

Step 2: Rewrite the equation in exponential form. Remember the rule: log_b(c) = a becomes b^a = c. Here, our base b is 5, our argument c is 125, and our exponent a is x. So, the equation becomes: 5^x = 125

Step 3: Solve for the variable. Now the question is much simpler: "What power of 5 gives us 125?" Let's test it out:

  • 5¹ = 5
  • 5² = 25
  • 5³ = 125 We found it! The exponent we need is 3. So, x = 3.
Example 2

Evaluating a Logarithm with a Fractional Solution

Problem: Evaluate log₆₄(4).

Step 1: Set the expression equal to a variable. log₆₄(4) = x

Step 2: Rewrite the equation in exponential form. The base is 64, the argument is 4. 64^x = 4

Step 3: Solve for the variable. This is where students often get stuck. How can raising a bigger number (64) to a power give you a smaller number (4)? This is a huge clue that the exponent must be a fraction—specifically, a root! We need to find a common base for both 64 and 4. The number 4 works perfectly. We know that 4³ = 64. Let's substitute that into our equation. (4³)^x = 4

Now, use the power-to-a-power exponent rule, which says (a^m)^n = a^(m*n). 4^(3x) = 4

Remember that 4 is the same as 4¹. 4^(3x) = 4¹

Since the bases are the same, the exponents must be equal. 3x = 1 x = 1/3

Example 3

Evaluating a Common Logarithm

Problem: Evaluate log(0.01).

Step 1: Identify the base and set the expression equal to a variable. There's no base written, so this is the common logarithm with base 10. log₁₀(0.01) = x

Step 2: Rewrite the equation in exponential form. 10^x = 0.01

Step 3: Solve for the variable. It's often easiest to convert decimals to fractions when dealing with powers of 10. 0.01 is one-hundredth, or 1/100. 10^x = 1/100

We know 10² = 100. To get 1/100, we need a negative exponent. 10⁻² = 1/10² = 1/100. So, x = -2.

Try it yourself

Time to try a couple on your own. Remember the process: set it equal to a variable, rewrite it, and solve.

Problem 1: Evaluate log₄(64)

Hint: Ask yourself, "4 to what power equals 64?"

Problem 2: Evaluate log(0.1)

Hint: What is the unwritten base here? And how can you write the decimal 0.1 as a fraction?

You've got this. Take your time and focus on that rewriting step.

Table for evaluating `log₄(64)` and `log(0.1)`.
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