Answer
$=\dfrac {1}{9}$
Work Step by Step
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,
$4^{-x} = 3^{-2}
\\4^{-x} = \dfrac{1}{3^2}
\\4^-x=\dfrac{1}{9}$
