Log Of Powers

Theorem: $\:\:log{(x^y)} = y\; log{(x)}$

Prerequisites:
Logarithm (definition)
Product of the powers property (proof)

Proof:
Let $x = a^z,\;$ where $a$ is the base of the log. Then by definition of log,

$\qquad\quad log_a{(x)} = z\qquad\qquad\qquad \qquad\;\;\;\cdots (1)$

$\begin{align}\text{Now, }\; log_a{(x^y)} &= log_a{\big(a^z\big)^y}\\
&= log_a{(a^{yz})}\qquad\qquad\quad\text{(by product of the power property)}\\
&= yz\qquad\qquad\qquad\qquad\text{(by definition of log)}\\
&= y\; log_a{(x)}\qquad\qquad\quad\text{(from $(1)$)}\end{align}$

Hence the result.

Recommended:
Log of product
Multiplication of log
Change of base property