Product Of The Powers

Theorem: $\;\;\big(a^{x}\big)^y = a^{\displaystyle (xy)}\quad\forall\; x, y \in \mathbb{R} \text{ and }  a > 0$

Prerequisites:
Exponentiation (definition)
Sum of the Powers Property (proof)
Continuity of exponential function

Proof:

Case I: When $y\in\mathbb{Z}$

$\qquad\quad\begin{align}\big(a^{x}\big)^y &= a^x.a^x.a^x....a^x\qquad\;\;\text{($y$ times)}\\ &= a^{\displaystyle (x+x+x...x)}\qquad\text{(by sum of powers property)}\\ &= a^{\displaystyle (xy)}\end{align}$

Case II: When $y\in\mathbb{Q}$

Let $y=p/q$, where $p, q\in\mathbb{Z}$

Let $\quad a^{\tfrac{p}{q}x} = b\qquad\qquad\qquad\cdots\text{(1)}$
$\therefore\quad\;\; \big(a^{\tfrac{x}{q}}\big)^p = b\qquad\qquad\qquad\text{(from case I)}\\ \Rightarrow\quad\;\; a^{\tfrac{x}{q}} = b^{\tfrac{1}{p}}\qquad\qquad\qquad\;\text{(by definition of exponentiation)}$

Raising both sides to the power $q$,

$\therefore\quad\;\; \big(a^{\tfrac{x}{q}}\big)^q = \big(b^{\tfrac{1}{p}}\big)^q\\ \Rightarrow\quad\;\; a^{\tfrac{x}{q}q} = b^{\tfrac{q}{p}}\qquad\qquad\qquad\text{(from case I)}\\ \Rightarrow\quad\;\; a^{x} = b^{\tfrac{q}{p}}$

Let $q/p=r$
$\therefore\quad\quad a^{x} = b^r\\ \Rightarrow\quad\;\; \big(a^x\big)^{\tfrac{1}{r}} = b\qquad\qquad\qquad\text{(by definition of exponentiation)}\\ \Rightarrow\quad\;\; \big(a^x\big)^{\tfrac{1}{r}} = b = a^{\tfrac{p}{q}x}\qquad\quad\!\!\text{(from $(1)$)}$

Case III: When $y\in\mathbb{R}$

If $y$ is irrational, then in any neighbourhood of $y$, there exists infinitely many rational numbers. Since at these rational numbers, the given equality holds, hence by the continuity of the exponential functions, the equality must also hold at $y$.

Therefore the result holds at all real values of $x$ and $y$.


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