Answer
$log_4(-2+\sqrt 7)\approx- 0.315$
Work Step by Step
Step 1. Let $u=4^x$, we have $u^2+4u-3=0 \longrightarrow u=\frac{-4\pm\sqrt {16+12}}{2}=-2\pm\sqrt 7$
Step 2. For $4^x=-2-\sqrt 7$, there is no real solution.
Step 3. For $4^x=-2+\sqrt 7$, we have $x=log_4(-2+\sqrt 7)\approx- 0.315$
