How to Prove $a^{\log_b c}=c^{\log_b a}$

For real logarithms, the identity

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

holds whenever $a>0$, $c>0$, $b>0$, and $b\ne 1$. The quickest proof is to rewrite both logarithms using natural logarithms. The resulting expressions have the same exponent because multiplication is commutative.

Proof using change of base

The change-of-base formula is

$$\log_b x=\frac{\ln x}{\ln b},$$

which applies when $x>0$, $b>0$, and $b\ne1$. Apply it to the exponent on the left:

$$\begin{aligned} a^{\log_b c} &=a^{\frac{\ln c}{\ln b}}\\ &=\exp\left(\ln a\,\frac{\ln c}{\ln b}\right)\\ &=\exp\left(\frac{\ln a\,\ln c}{\ln b}\right). \end{aligned}$$

Do the same on the right:

$$\begin{aligned} c^{\log_b a} &=c^{\frac{\ln a}{\ln b}}\\ &=\exp\left(\ln c\,\frac{\ln a}{\ln b}\right)\\ &=\exp\left(\frac{\ln c\,\ln a}{\ln b}\right). \end{aligned}$$

Since $\ln a\,\ln c=\ln c\,\ln a$, the exponents are identical. Therefore,

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

Another proof using a common base

Set $u=\log_b a$ and $v=\log_b c$. By the definition of logarithm, $a=b^u$ and $c=b^v$. Then

$$a^{\log_b c}=a^v=(b^u)^v=b^{uv},$$

while

$$c^{\log_b a}=c^u=(b^v)^u=b^{vu}.$$

Because $uv=vu$, both sides equal $b^{uv}$. This proof highlights the same symmetry without converting to natural logarithms.

Worked example

Take $a=4$, $b=2$, and $c=8$. The logarithms are exact integers:

$$\log_2 8=3,\qquad \log_2 4=2.$$

Thus

$$4^{\log_2 8}=4^3=64,$$

and

$$8^{\log_2 4}=8^2=64.$$

The example illustrates the identity, while the proof shows why it works for every admissible positive $a$, $b$, and $c$.

Why the conditions matter

For a real logarithm $\log_b x$, its argument must be positive and its base must be positive and different from $1$. Here the arguments are $a$ and $c$, so both must be positive; the shared base $b$ must satisfy $b>0$ and $b\ne1$. These conditions also make the real powers in the identity well-defined under the usual definition $x^r=\exp(r\ln x)$ for $x>0$.

There is no requirement that $a$, $b$, and $c$ be distinct. For example, $a$ may equal $b$, or $a$ may equal $c$, as long as the domain conditions still hold.

Common mistakes

• Forgetting the logarithm base restrictions: $\log_1 x$ is undefined, and real logarithms with a nonpositive base are not defined in the standard sense.
• Allowing a nonpositive argument: If $a\le0$ or $c\le0$, one of the logarithms is not a real number.
• Treating the equality as a special exponent rule: The key step is not a rule that freely swaps a base and exponent. It works here because change of base makes the exponent involve the symmetric product $\ln a\,\ln c$.
• Changing only one side: To compare the expressions reliably, rewrite both logarithms or use the common-base substitution on both sides.