#### Answer

$\dfrac{1}{49}$

#### Work Step by Step

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, $4^{-2x}=(4^{-x})^{2} =(\dfrac{1}{7})^2=\dfrac{1}{49}$