## Precalculus (10th Edition)

$\dfrac {1}{27}$
$a^{m\times n}=\left( a^{m}\right) ^{n} \\\Rightarrow 5^{3x}=\left( 5^{x}\right) ^{3} \\5^{3x}=\left( 5^{-x}\right) ^{-3}$ Since $5^{-x}=3$, then $5^{3x}=(5^{-x})^3 \\5^{3x}=3^{-3}$ Using the rule $a^{-m} = \dfrac{1}{a^m}$ gives: $\\5^{3x}=\dfrac {1}{3^3} \\5^{3x}=\dfrac{1}{27}$