The Inverse of a Logarithmic Function

Idea

This is the previous lesson run backwards. There, $x$ was trapped in a power and a logarithm freed it. Here $x$ is trapped inside a logarithm, and it takes an exponential to free it.

Students reverse these two. Which function you reach for depends on where $x$ is stuck, not on what the question is about.

Rule

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

$a^{\log_a x} = x$

How it is used

1. Where is x stuck

$x$ appears	apply	leaves
inside a power, $a^{\square} = k$	$\log_a$	$\square = \log_a k$
inside a logarithm, $\log_a \square = k$	$a$ to the power	$\square = a^k$

Both rows say the same thing: use the function that undoes the one holding $x$ .

2. Make the logarithm stand alone first

The exponential step only works on a bare logarithm. Clear everything around it before raising the base.

3\log_2 x - 4 = y \;\Rightarrow\; \log_2 x = \frac{y + 4}{3} \;\Rightarrow\; x = 2^{(y + 4)/3}

Raising the base while a coefficient is still attached is the mistake here. $2^{3\log_2 x}$ is not $x$ .

3. It releases the whole argument

Whatever sits inside the logarithm comes out in one piece.

\log_a(x + k) = y \;\Rightarrow\; x + k = a^y

The $k$ then comes off by ordinary algebra, as a separate step.

4. Domain and range change places again

	domain	range
$f(x) = \log_a x$	$x > 0$	every real $y$
$f^{-1}(x) = a^x$	every real $x$	$y > 0$

A logarithm accepts only positive numbers and returns anything. Its inverse accepts anything and returns only positive numbers.

Exponential and logarithm

Try this. Compare bases 2 and 0.5: growth becomes decay. Move the input to 1; the logarithm is zero for either base.

BasePositive input x0.5

Base 3 · Blue: y = 3ˣ · Green: y = log_b(x) · Dashed: y = x. At x = 0.5, y = -0.631.

Worked example

Find the inverse of $f(x) = 3\log_2(x + 1) - 4$ .

Remember

$a^{\log_a x} = x$ , so raising the base undoes a logarithm clear everything around the logarithm before raising the base an inverse is written in $x$ , so swap the letters at the end

Write $y$ for $f(x)$ .

y = 3\log_2(x + 1) - 4

Add the constant, then divide by the coefficient. Only now is the logarithm alone.

\frac{y + 4}{3} = \log_2(x + 1)

Raise $2$ to both sides. It releases the whole bracket, not $x$ alone.

x + 1 = 2^{(y + 4)/3}

Subtract $1$ , then swap the letters.

f^{-1}(x) = 2^{(x + 4)/3} - 1

Explore

Drag the probe along the green logarithm curve and watch its partner on the blue exponential.

Notice where each curve lives. The logarithm exists only to the right of the $y$ -axis, because its argument must be positive. Its inverse exists everywhere but stays above the $x$ -axis. That is the same restriction, seen from the other side.