Answer
$log_3(-2+\sqrt 7)\approx-0.398$
Work Step by Step
Step 1. Let $u=3^x$, we have $u^2+4u-3=0$, thus $u=\frac{-4\pm\sqrt {16+12}}{2}=-2\pm\sqrt 7$
Step 2. For $3^x=-2+\sqrt 7$, we get $x=log_3(-2+\sqrt 7)\approx-0.398$
Step 3. For $3^x=-2-\sqrt 7$, there is no real solution.