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AP Precalculus · Lesson 2.4

Logarithms as Inverses

How it works

A logarithm answers the exponent question. For a positive base other than one, the logarithm undoes exponentiation. Real logarithms require positive arguments. The inverse exchanges the original function domain and range.

log⁡by=x⟺bx=y(b>0, b≠1, y>0)\log_b y=x\quad\Longleftrightarrow\quad b^x=y\quad(b>0,\ b\ne1,\ y>0)

Worked examples

Example 1

Evaluate log⁡28\displaystyle \log_2 8.

Since 23=8\displaystyle 2^3=8, the value is 3\displaystyle 3.

Example 2

Find the inverse of f(x)=3x\displaystyle f(x)=3^x.

Interchange input and output and solve: f−1(x)=log⁡3x\displaystyle f^{-1}(x)=\log_3x for x>0\displaystyle x>0.

Problems

Problem 2.4.1

Evaluate log⁡5(1/25)\displaystyle \log_5(1/25).

Show worked solution

Since 5−2=1/25\displaystyle 5^{-2}=1/25, the logarithm is −2\displaystyle -2.

Problem 2.4.2

State the domain of g(x)=ln⁡(x−4)\displaystyle g(x)=\ln(x-4).

Show worked solution

The argument must be positive: x−4>0\displaystyle x-4>0, so the domain is (4,∞)\displaystyle (4,\infty).

Problem 2.4.3

Solve log⁡3x=4\displaystyle \log_3x=4.

Show worked solution

Convert to exponential form: x=34=81\displaystyle x=3^4=81.

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