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

$x \approx 1.161$
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: $x \approx 1.161$