Exponent Combination Laws/Difference of Powers

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Theorem

Let $a \in \R_+$ be a positive real number.

Let $x, y \in \R$ be real numbers.

Let $a^x$ be defined as $a$ to the power of $x$.

Then:

$\dfrac{a^x}{a^y} = a^{x-y}$

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \frac{a^x}{a^y}$	$=$	$\displaystyle a^x \left({\frac 1 {a^y} }\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \left({a^x}\right) \left({a^{-y} }\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Exponent Combination Laws: Negative Power
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle a^{x-y}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Exponent Combination Laws: Sum of Powers

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac{a^x}{a^y}\)	\(=\)	\(\displaystyle a^x \left({\frac 1 {a^y} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({a^x}\right) \left({a^{-y} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Exponent Combination Laws: Negative Power
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a^{x-y}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Exponent Combination Laws: Sum of Powers