## Precalculus (10th Edition)

$=\dfrac {1}{9}$
RECALL: $a^{m\times n}=\left( a^{m}\right) ^{n}$ Thus, $4^{-x}=\left( 4^{x}\right) ^{-1}$ Note that $4=2^2$, so : $4^{-x}=\left(( 2^2)^x \right)^{-1} \\4^{-x}=\left( 2^{2x}\right)^{-1} \\4^{-x}=2^{(-2x)} \\4^{-x}=(2^x)^{-2}$ Since $2^x=3$, then $4^{-x}=3^{-2}$ RECALL: $a^{-m} = \dfrac{1}{a^m}$ Therefore, $4^{-x} = 3^{-2} \\4^{-x} = \dfrac{1}{3^2} \\4^-x=\dfrac{1}{9}$