## College Algebra (10th Edition)

$\dfrac{1}{9}$
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: $=\dfrac{1}{3^2} \\=\dfrac{1}{9}$