ACT Math: Logarithm

Logarithm questions on the ACT are relatively rare, usually just a few per test, but they're highly learnable once you see logarithms for what they really are: the inverse of exponentials. This free, complete lesson covers the logarithm-exponential relationship, properties of logarithms (the product, quotient, and power rules), solving logarithmic and exponential equations, and natural logarithms with the change of base formula. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback, in the same four-answer-choice, calculator-allowed format as the current ACT. Everything here is free, and you can keep practicing afterward with the full ACT Math Question Bank linked below.

When a problem just writes "log x" with no base shown, it means base 10. When a problem writes "ln x," it means the natural logarithm, base e. A calculator's log button computes base-10 logarithms directly.

Practice Logarithm Free Explore the Full ACT Math Qbank

1. The Logarithm-Exponential Relationship

A logarithm answers the question, "what power do I raise the base to, in order to get this number?" Logarithms and exponentials are inverse operations of each other:

log_b(x) = y is equivalent to b^y = x

Worked Example: Evaluate log₅(125).

Ask: "5 raised to what power gives 125?" Since 5³ = 125:

log₅(125) = 3

1. Evaluate log₂(32).

Explanation: 2⁵ = 32, so log₂(32) = 5.

2. Evaluate log₃(81).

Explanation: 3⁴ = 81, so log₃(81) = 4.

3. Evaluate log(1000). (Base 10 is implied.)

Explanation: 10³ = 1,000, so log(1,000) = 3.

4. Rewrite log₄(64) = 3 in exponential form.

Explanation: The base (4) raised to the result (3) equals the original number (64): 4³ = 64.

5. Rewrite 2⁶ = 64 in logarithmic form.

Explanation: The base of the exponential (2) becomes the base of the log, and the exponent (6) becomes the result: log₂(64) = 6.

2. Properties of Logarithms

Rule	Formula
Product rule	log_b(xy) = log_b(x) + log_b(y)
Quotient rule	log_b(xy) = log_b(x) − log_b(y)
Power rule	log_b(xⁿ) = n · log_b(x)

Worked Example: Simplify log₂(8) + log₂(4).

Using the product rule: log₂(8) + log₂(4) = log₂(8 × 4) = log₂(32).

2⁵ = 32, so log₂(32) = 5

1. Simplify: log₃(9) + log₃(27)

Explanation: log₃(9) = 2 and log₃(27) = 3, so their sum is 5.

2. Simplify: log₂(64) − log₂(8)

Explanation: log₂(64) = 6 and log₂(8) = 3, so their difference is 3. Alternatively, by the quotient rule: log₂(64 ÷ 8) = log₂(8) = 3.

3. Simplify: log₅(25³)

Explanation: By the power rule: log₅(25³) = 3 · log₅(25) = 3 × 2 = 6.

4. Given log(2) ≈ 0.301 and log(3) ≈ 0.477, approximate log(6).

Explanation: log(6) = log(2 × 3) = log(2) + log(3) ≈ 0.301 + 0.477 = 0.778.

5. Which properties are used to simplify log(x²y) as 2 log(x) + log(y)?

Explanation: The product rule splits log(x²y) into log(x²) + log(y), and the power rule then rewrites log(x²) as 2 log(x).

3. Solving Logarithmic and Exponential Equations

To solve a logarithmic equation, rewrite it in exponential form. To solve an exponential equation, either rewrite both sides with the same base (and set the exponents equal) or take a logarithm of both sides. Always check logarithmic solutions against the original equation, since a logarithm's argument must be positive.

Worked Example: Solve log₂(x) = 5.

x = 2⁵ = 32

1. Solve: log₄(x) = 3

Explanation: x = 4³ = 64.

2. Solve: log(x) = −2 (base 10)

Explanation: x = 10⁻² = 1100 = 0.01.

3. Solve: 3^x = 81

Explanation: 81 = 3⁴, so x = 4.

4. Solve: 2^x+1 = 32

Explanation: 32 = 2⁵, so x + 1 = 5, giving x = 4.

5. Solve: log₃(x + 2) + log₃(x − 4) = 3

Explanation: Using the product rule: log₃[(x+2)(x−4)] = 3, so (x+2)(x−4) = 27. Expanding: x² − 2x − 35 = 0, which factors as (x − 7)(x + 5) = 0, giving x = 7 or x = −5. Since x − 4 must be positive for the original logarithm to be defined, x = −5 is rejected. Only x = 7 works.

4. Natural Logarithms and Change of Base

A natural logarithm, written ln(x), is a logarithm with base e (an irrational constant, e ≈ 2.718). Since ln and e^x are inverse functions, ln(e^x) = x for any x. The change of base formula lets you compute any logarithm using a calculator's base-10 (or natural) log key:

log_b(a) = log(a)log(b)

Worked Example: Evaluate ln(e⁴).

ln(e⁴) = 4

1. Evaluate ln(e⁷).

Explanation: Since ln and e^x are inverse functions, ln(e⁷) = 7.

2. Evaluate ln(1).

Explanation: Since e⁰ = 1, ln(1) = 0. This holds for logarithms of any base: log_b(1) = 0.

3. Solve: ln(x) = 3

Explanation: ln(x) = 3 means log_e(x) = 3, so x = e³.

4. Use the change of base formula to rewrite log₇(20) using base-10 logarithms.

Explanation: The change of base formula puts the original argument (20) on top and the original base (7) on the bottom: log(20)log(7).

5. A population grows according to P(t) = 500e^0.03t. After how many years will the exponent equal 1.5 (that is, 0.03t = 1.5)?

Explanation: 0.03t = 1.5, so t = 1.5 ÷ 0.03 = 50.

Common Mistakes to Avoid

Forgetting that a logarithm's argument must be positive. log_b(negative number) is undefined, always check that a solved value keeps every logarithm's argument positive in the original equation.
Confusing the product rule with the power rule. log_b(xy) = log_b(x) + log_b(y), a sum, while log_b(xⁿ) = n · log_b(x), a product with the exponent pulled out front. They apply to different situations.
Treating log(x) + log(y) as log(x + y). The product rule turns a sum of logs into the log of a product, not the log of a sum, this is a very common algebra slip.
Mixing up which number is the base and which is the argument when converting between forms. In log_b(x) = y, b is always the base, x is the argument, and y is the result (the exponent).
Forgetting that ln always means base e, not base 10. "log" with no subscript defaults to base 10; "ln" always means the natural logarithm, base e. Mixing these up leads to using the wrong calculator button.

Frequently Asked Questions

Why are logarithms and exponentials called inverse functions?

An exponential function takes a power and produces a number; a logarithm takes that number and gives back the power. Applying one after the other, in either order, returns you to where you started, which is exactly what it means for two functions to be inverses of each other.

Do I need to memorize the change of base formula?

Yes, it's essential for evaluating a logarithm with an unusual base (like log₇(20)) using a standard calculator, which typically only has buttons for base-10 (log) and base-e (ln) logarithms.

Why do I need to check solutions to logarithmic equations?

Solving a logarithmic equation algebraically can sometimes produce a value that makes one of the original logarithm's arguments zero or negative, which is mathematically undefined. Always substitute your solution back into the original equation to confirm it's valid.

Where can I practice more problems like these?

The Logarithm quizzes in the ACT Math Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested on the ACT Math section.