#### Answer

$x \approx 1.161$

#### Work Step by Step

We wish to substitute $t=4^{x}$, so the equation becomes:
$a -(10)(\dfrac{1}{t}) -3=0 \implies a^{2}-3t-10=0 $
This gives a quadratic equation, whose factors are: $(t-5)(t+2)=0$
Use the zero factor property to obtain:
$ t-5 =0 \implies t=5$ and $t+2=0 \implies t=-2$
But $t=4^{x}=-2$ cannot be a solution as $4^{x}$ can never be negative. So, we will consider $t=4^{x}=5$
Therefore, $4^{x}=5$
or, $x= \log_4 5 \approx 1.161$
Thus, our answer is: $x \approx 1.161$