## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{1}{49}$
In order to solve the given problem, we will use the following two rules as follows: $(a) a^{-p}=\dfrac{1}{a^p} \\ (b) a^{pq}=(a^p)^q$ We will use rule-(a) as: $4^{-2x}=\dfrac{1}{4^{2x}}$ Now, we will use rule-(b) as: $4^{-2x}=\dfrac{1}{(4^{x})^2}$ When $4^x=7$, then we simplify the expression as: $4^{-x}=\dfrac{1}{7}$ Therefore, $4^{-2x}=(4^{-x})^{2} =(\dfrac{1}{7})^2=\dfrac{1}{49}$