Exponent Combination Laws/Negative Power of Quotient

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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

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

$\blacksquare$