Why Are Exponential and Logarithmic Functions Inverses of Each Other?

Exponential and logarithmic functions are inverses because each one reverses the operation performed by the other. If exponentiation maps an input $x$ to $a^x$, then the logarithm base $a$ maps $a^x$ back to $x$.

This is part of Mathematics: Analysis and Approaches, SL 2.9: Exponential and logarithmic functions, and may be assessed on Paper 1 or Paper 2.

An inverse function reverses a function so that both compositions return the original input. Let

$$f(x)=a^x \quad \text{and} \quad g(x)=\log_a x,$$

where $a>0$ and $a\neq 1$. By the definition of a logarithm,

$$a^x=y \iff \log_a y=x.$$

Therefore,

$$g(f(x))=\log_a(a^x)=x$$

and

$$f(g(x))=a^{\log_a x}=x, \qquad x>0.$$

Both compositions give the identity function, confirming that $g=f^{-1}$. This works because $a^x$ is one-to-one when $a>0$ and , so every positive output corresponds to exactly one real input. The exponential function never produces zero or a negative value, which explains why the logarithm's domain is restricted to positive real numbers.

The inverse relationship also explains why the domain and range must be exchanged. For $f$, every real input is allowed, but every output is positive. For $g$, only positive inputs are allowed, while every real number can occur as an output. These matching sets make both compositions valid on their appropriate domains. If $a$ equals $1$, the exponential function is constant and therefore has no inverse.

For example, using base $2$,

$$2^3=8 \iff \log_2 8=3.$$

The exponential function sends $3$ to $8$, while the logarithmic function sends $8$ back to $3$. Similarly, solving $3^x=20$ means applying the inverse operation:

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

Their domains and ranges are exchanged, and corresponding points swap coordinates. Consequently, their graphs are reflections of each other in $y=x$, as expected for inverse functions.

A common misconception is that an inverse function is a reciprocal. In fact, $f^{-1}(x)=\log_a x$, not $1/a^x$.

Exam technique: To justify that two functions are inverses, show both compositions equal $x$, or interchange $x$ and $y$ and rearrange. State $a>0$, $a\neq1$, and the logarithmic restriction .