Exponent Combination Laws/Negative Power of Quotient

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Theorem

Let $a, b \in \R_{>0}$ be (strictly) 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

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

$\blacksquare$