# Exponent Combination Laws/Negative Power of Quotient

## Theorem

Let $a, b \in \R_+$ be positive real numbers.

Let $x \in \R$ be a real number.

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

Then:

$\left({\dfrac a b}\right)^{-x} = \left({\dfrac b a}\right)^x$

## Proof

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \left({\frac a b}\right) ^{-x}$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \left({\frac 1 {\left({\frac a b}\right)} }\right)^x$$ $$\displaystyle$$ $$\displaystyle$$ Exponent Combination Laws/Negative Power $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \left({\frac b a}\right)^x$$ $$\displaystyle$$ $$\displaystyle$$

$\blacksquare$