When a Logarithm Is the Exponent

Idea

A logarithm is an exponent, so there is nothing strange about finding one sitting in an exponent slot. $2^{\log_2 5}$ reads: raise $2$ to the power that turns $2$ into $5$ .

Written that way the answer is already visible. It is $5$ . The work in this lesson is getting every expression into that shape.

Rule

$a^{\log_a x} = x$

$a^{m \log_a x} = x^m$

$a^{\log_b c} = c^{\log_b a}$

How it is used

1. A multiplier in the exponent becomes a power

Split the exponent before cancelling, and the known pair appears.

a^{m \log_a x} = \left(a^{\log_a x}\right)^m = x^m

3^{2 \log_3 4} = 4^2 = 16,

e^{2 \ln 3} = 3^2 = 9,

5^{3 \log_5 2} = 2^3 = 8

2. Making the two bases match

If the base of the power is a power of the base of the logarithm, rewrite it and the multiplier drops into the exponent.

4^{\log_2 3} = (2^2)^{\log_2 3} = 2^{2 \log_2 3} = 3^2 = 9

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.

Expression	Base rewritten	Value
$9^{\log_3 5}$	$3^{2 \log_3 5}$	$25$
$8^{\log_2 5}$	$2^{3 \log_2 5}$	$125$
$25^{\log_5 3}$	$5^{2 \log_5 3}$	$9$
$\left(\dfrac{1}{9}\right)^{\log_3 2}$	$3^{-2 \log_3 2}$	$\dfrac{1}{4}$

Spotting that $8$ is $2^3$ is the whole skill. If no common base exists, nothing cancels.

3. Extra terms in the exponent

A sum in an exponent is a product of powers, so peel the ordinary part off first.

2^{\log_2 5 + 3} = 2^{\log_2 5} \cdot 2^3 = 5 \cdot 8 = 40

5^{\log_5 2 - 1} = 5^{\log_5 2} \cdot 5^{-1} = \frac{2}{5}

4. Swapping the two ends

Take $\log_b$ of each side of $a^{\log_b c} = c^{\log_b a}$ and use the power rule:

\log_b\left(a^{\log_b c}\right) = \log_b c \cdot \log_b a,

\log_b\left(c^{\log_b a}\right) = \log_b a \cdot \log_b c

The two products are the same number, and a logarithm never gives one value to two different arguments, so the powers themselves are equal. It turns an awkward expression into an easy one:

81^{\log_3 2} = 2^{\log_3 81} = 2^4 = 16

Worked example

Evaluate $8^{\log_2 5} + 2^{\log_2 3 + 2}$ .

Remember

$a^{\log_a x} = x$ only when the two bases match $(a^m)^n = a^{mn}$ , so a multiplier in the exponent is a power outside $a^{m+n} = a^m \cdot a^n$

The first term has base $8$ and a logarithm in base $2$ . Rewrite $8$ .

8^{\log_2 5} = (2^3)^{\log_2 5} = 2^{3 \log_2 5}

Pull the $3$ back out as an outside power, and the matching pair cancels.

2^{3 \log_2 5} = \left(2^{\log_2 5}\right)^3 = 5^3 = 125

The second exponent is a sum, so the power splits into a product.

2^{\log_2 3 + 2} = 2^{\log_2 3} \cdot 2^2 = 3 \cdot 4 = 12

Add the two terms.

125 + 12 = 137

Explore

The height of the curve above $x$ is the exponent that the base needs to reach $x$ . Set the base to $3$ and read the height above $x = 9$ : it is $2$ . Feeding that height back as an exponent, $3^2 = 9$ , returns the argument you started from. That round trip is $a^{\log_a x} = x$ .

Drag the probe to $x = 2$ and read the height, roughly $0.63$ . Doubling it and using it as an exponent gives $3^{1.26} \approx 4$ , which is $2^2$ , the same thing $a^{2 \log_a x} = x^2$ predicts. The curve supplies the exponent; the exponential spends it.