Answer
$\dfrac{1}{9}$
Work Step by Step
Note that $4=2^2$.
Thus, $4^{-x} = (2^2)^{-x}$
RECALL:
(1) $a^{-m} = \dfrac{1}{a^m}$
(2) $a^{mn}= (a^m)^n$
Use rule (2) above to obtain:
$4^{-x} = 2^{2(-x)} = 2^{-2x}$
Use rule (1) above to obtain:
$2^{-2x} = \dfrac{1}{2^{2x}}$
Use rule (2) above to obtain:
$\dfrac{1}{2^{2x}}=\dfrac{1}{(2^x)^2}$
Thus,
$4^{-x} = \dfrac{1}{(2^x)^2}$
With $2^x=3$, the expression above simplifies to:
$=\dfrac{1}{3^2}
\\=\dfrac{1}{9}$
