# Proof

Prove if $x$ and $b$ are positive real values with $b \neq 1$, and $n$ is any real value, then

$$\displaystyle{\log_b x^n = n\log_b x}$$

First note that we can evaluate $\log_b x$, since $x$ and $b$ are positive real values with $b \neq 1$. Let us denote this value by $m$, so

$$\log_b x = m$$

Expressing the above in exponential form, we then have $$b^m = x$$

Now the rest of the proof follows quickly, as shown below. Notice how the proof hinges on the fact that $(b^m)^n = b^{mn}$, one of our laws of exponents.

$$\begin{array}{rcl} \log_b x^n &=& \log_b (b^m)^n \\ &=& \log_b b^{mn}\\ &=& mn\\ &=& n\log_b x \end{array}$$