If 2⁵ = 32, then the exponent 5 is the answer to the question: "2 to what power gives 32?" A logarithm is exactly that question written as a function. Algebra II extends exponential functions to logarithmic ones, and the payoff is a way to solve equations where the unknown sits in an exponent, which ordinary algebra cannot reach.
The definition
For a positive base b not equal to 1 and a positive number y:
log_b(y) = x means bˣ = y.
Read it as "the logarithm base b of y is the exponent you put on b to get y." The two forms say the same thing, and moving between them is the most useful skill in this topic.
- log₂(8) = 3, because 2³ = 8.
- log₁₀(1000) = 3, because 10³ = 1000.
- log₅(1/25) = −2, because 5⁻² = 1/25.
- log_b(1) = 0 for every base, because b⁰ = 1.
- log_b(b) = 1, because b¹ = b.
Two bases get special buttons on a calculator. The common logarithm, written log, has base 10. The natural logarithm, written ln, has base e, where e ≈ 2.718.
Logarithms undo exponentials
The function y = bˣ and the function y = log_b(x) are inverses. That gives two identities: log_b(bˣ) = x and b^(log_b x) = x. The graph of one is the reflection of the other across the line y = x. Because bˣ is always positive, log_b(x) is defined only for x > 0.
The properties
These follow from the laws of exponents. For positive M and N:
- log_b(MN) = log_b(M) + log_b(N)
- log_b(M ÷ N) = log_b(M) − log_b(N)
- log_b(Mᵖ) = p · log_b(M)
The third one is the key to solving exponential equations, because it brings an exponent down to the ground. There is also the change-of-base idea: log_b(y) = ln(y) ÷ ln(b), so any logarithm can be evaluated with a calculator's log or ln key.
Solving exponential equations
Worked example 1. Solve 2ˣ = 20. Take the logarithm of both sides (base 10 here):
x · log 2 = log 20, so x = log 20 ÷ log 2 ≈ 1.3010 ÷ 0.3010 ≈ 4.32.
Check: 2⁴ = 16 and 2⁵ = 32, and 20 is between them, so x between 4 and 5 makes sense.
Worked example 2. Solve 3 · 5ˣ = 60. First isolate the exponential: 5ˣ = 20. Then x = ln 20 ÷ ln 5 ≈ 2.9957 ÷ 1.6094 ≈ 1.86. Check: 5¹ = 5 and 5² = 25, and 20 is between them.
Worked example 3: doubling money. An account earns 4% interest a year, compounded annually, so the balance after t years is 1000 · (1.04)ᵗ. When does it reach $2000? Solve 1.04ᵗ = 2: t = ln 2 ÷ ln 1.04 ≈ 0.6931 ÷ 0.03922 ≈ 17.7 years. Check: 1.04¹⁷ ≈ 1.948 and 1.04¹⁸ ≈ 2.026, so it passes 2 during year 18.
One more case: if the unknown is in the exponent and both sides share a base, you do not need logarithms. Solve 4ˣ = 32 by writing both sides as powers of 2: (2²)ˣ = 2⁵, so 2x = 5 and x = 2.5. Check: 4^2.5 = 2⁵ = 32. Logarithms are for cases like 2ˣ = 20, where no such match exists.
Solving logarithmic equations
To solve an equation with a logarithm in it, combine logarithms with the properties, then rewrite in exponential form.
Worked example 4. Solve log₂(x) + log₂(x − 2) = 3.
- Combine: log₂(x(x − 2)) = 3.
- Rewrite in exponential form: x(x − 2) = 2³ = 8.
- Expand: x² − 2x − 8 = 0, which factors as (x − 4)(x + 2) = 0.
- So x = 4 or x = −2.
Now check the domain. If x = −2, then log₂(−2) is not defined, so reject it. For x = 4: log₂(4) + log₂(2) = 2 + 1 = 3. The only solution is x = 4.
Common mistakes
- Splitting a logarithm of a sum. log(a + b) is not log(a) + log(b). Only products turn into sums.
- Taking the logarithm before isolating the exponential. In 3 · 5ˣ = 60, divide by 3 first. Otherwise you must handle log(3 · 5ˣ) = log 3 + x log 5, which is correct but easier to slip on.
- Forgetting the domain. Always check that the arguments of every logarithm in the original equation are positive.
- Writing ln 20 ÷ ln 5 as ln(20 ÷ 5). The quotient of two logarithms is not the logarithm of a quotient.
- Mixing bases. Use the same base on both sides of the equation. Either log or ln works, as long as you use it throughout.
Practice Algebra II questions with an explanation for every answer, or read the guide to California Algebra II practice.
Source: the California Common Core State Standards for Mathematics, F-LE.4, F-BF.5.