Power Of The Product

Theorem: $\;(ab)^x = a^x.b^x\;\; \forall\; a,b > 0 \;\text{ and }\; x\in\mathbb{R}$

Prerequisites:
Log of product property (proof)
Log of powers property (proof)

Proof:
Taking log of $(ab)^x$:

$\qquad\quad\begin{align} \log{(ab)^x} &= x\log{(ab)}\qquad\;\;\qquad\qquad\qquad\text{(by log of power property)}\\ &= x(\log{a} + \log{b})\qquad\qquad\qquad\text{(by log of product property)}\\ &= x\log{a} + x\log{b}\\ &= \log{a^x} + \log{b^x}\qquad\;\qquad\qquad\text{(by log of power property)}\\ &= \log{a^x.b^x}\qquad\qquad\;\;\qquad\qquad\text{(by log of product property)}\end{align}$

Taking antilog on both sides:

$\qquad\quad(ab)^x = a^x.b^x$

Hence the result


Recommended:
Sum of the powers
Product of the powers
Log of product
Log of powers