Answer
$x \approx -0.315$
Work Step by Step
Re-write as:
$ 16^{x}=(4^{2})^{x}=(4^{x})^{2}$ and $4^{x+1}=4^{x} \cdot 4$
We wish to substitute $a=4^{x}$, so the equation becomes:
$a^{2}+4a-3=0 $
Use the quadratic formula to find the roots:
$a=\displaystyle \frac{-4\pm\sqrt{(4^{2})-4(1)(-3)}}{2(1)}=\frac{-4\pm 2\sqrt{7}}{2}=-2\pm\sqrt{7}$
Since, $a=4^{x}$ is a positive term, we need to discard the negative value of $a$.
$a=-2+\sqrt{7}$
Therefore, $4^{x}=-2+\sqrt{7} $
or, $x=\log_{4}(-2+\sqrt{7})\approx-0.315$
Thus, the solution set is $x \approx -0.315$
