Swapping the Base and the Exponent

Idea

An exponent inside a logarithm has two places to sit: up in the argument, or down in the base. From either place it comes out to the front as a multiplier.

The only difference is which way up it arrives. From the argument it comes out as it stands; from the base it comes out inverted.

Rule

$\log_a x^m = m \log_a x$

$\log_{a^n} x = \dfrac{1}{n} \log_a x$

$\log_{a^n} x^m = \dfrac{m}{n} \log_a x$

How it is used

1. Why the exponent walks out

Let $k = \log_a x$ , so $a^k = x$ . Raise both sides to the power $m$ :

x^m = (a^k)^m = a^{km} \;\Rightarrow\; \log_a x^m = km = m \log_a x

Nothing was assumed. A power of a power multiplies exponents, and the logarithm is reading that exponent back.

\log_2 8^5 = 5 \log_2 8 = 5 \cdot 3 = 15,

\log_3 9^4 = 4 \log_3 9 = 4 \cdot 2 = 8

2. From the base it comes out inverted

Let $y = \log_{a^n} x$ , so $(a^n)^y = a^{ny} = x$ . Then $ny = \log_a x$ , and dividing by $n$ gives the second rule.

Logarithm	Base as a power	Value
$\log_4 8$	$\log_{2^2} 2^3$	$\dfrac{3}{2}$
$\log_9 27$	$\log_{3^2} 3^3$	$\dfrac{3}{2}$
$\log_8 4$	$\log_{2^3} 2^2$	$\dfrac{2}{3}$
$\log_{1/2} 8$	$\log_{2^{-1}} 2^3$	$-3$
$\log_{1/3} 9$	$\log_{3^{-1}} 3^2$	$-2$

A base below $1$ is a negative power of the base above it, which is where the minus signs come from.

3. Both exponents at once

Write base and argument as powers of one common number, then divide the argument exponent by the base exponent.

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 2 · Blue: y = 2ˣ · Green: y = log_b(x) · Dashed: y = x. At x = 0.5, y = -1.

Expression	As powers	Value
$\log_8 32$	$\log_{2^3} 2^5$	$\dfrac{5}{3}$
$\log_{27} 81$	$\log_{3^3} 3^4$	$\dfrac{4}{3}$
$\log_{16} 8$	$\log_{2^4} 2^3$	$\dfrac{3}{4}$
$\log_9 \sqrt{3}$	$\log_{3^2} 3^{1/2}$	$\dfrac{1}{4}$
$\log_{1/4} 8$	$\log_{2^{-2}} 2^3$	$-\dfrac{3}{2}$

Check any row by putting the base back: $8^{5/3} = 2^5 = 32$ .

4. What the rule does not say

$\log_a x^m$ and $(\log_a x)^m$ are different expressions. The exponent belongs to the argument in the first and to the whole logarithm in the second.

\log_2 4^3 = \log_2 64 = 6,

(\log_2 4)^3 = 2^3 = 8

One honest caveat: $\log_a x^2$ still makes sense for negative $x$ while $2 \log_a x$ does not, so the careful version is $\log_a x^2 = 2 \log_a |x|$ .

Worked example

Evaluate $\log_8 32 + \log_2 4^5$ .

Remember

$\log_a x^m = m \log_a x$ , the argument exponent comes out front $\log_{a^n} x = \dfrac{1}{n} \log_a x$ , the base exponent comes out inverted $\log_a a^n = n$ , so a matching base and argument read straight off

In the first term neither $8$ nor $32$ is the other's power, so write both in base $2$ .

\log_8 32 = \log_{2^3} 2^5

The base exponent $3$ divides and the argument exponent $5$ multiplies.

\log_{2^3} 2^5 = \frac{5}{3} \log_2 2 = \frac{5}{3}

In the second term only the argument carries an exponent, so it comes straight out.

\log_2 4^5 = 5 \log_2 4 = 5 \cdot 2 = 10

Add the two values over a common denominator.

\frac{5}{3} + 10 = \frac{5}{3} + \frac{30}{3} = \frac{35}{3}

Explore

Slide the base and watch how far the curve has to travel to reach a height of $1$ . That distance is the base itself, and squaring the base doubles the run needed for the same height. That doubling is exactly what $\log_{a^2} x = \frac{1}{2} \log_a x$ says: every value is halved.

Set the base to $2$ and read the height above $x = 8$ ; it is $3$ . Now set it to $4$ and read the same place: $\frac{3}{2}$ . The curve did not change shape, only scale, because raising the base to a power multiplies the whole logarithm by a constant.