Answer
$x =$ {$-0.876, 0.876 $}
Work Step by Step
Re-write as: $9^{x}=(3^{2})^{x}=(3^{x})^{2}$ and $3^{x+1}=3^{x} \cdot 4$
We wish to substitute $a=3^{x}$, so the equation becomes:
$a^{2}-3a+1=0 $
Use the quadratic formula to find roots.
$a=\displaystyle \frac{3 \pm\sqrt{(-3^{2})-4(1)(-3)}}{2(1)}=\frac{3 \pm \sqrt{5}}{2}$
Therefore, $3^{x}=\dfrac{3 \pm \sqrt{5}}{2}$
So, $x=\log_{3}\dfrac{3 + \sqrt{5}}{2} \approx 0.876$ and $x= \log_{3}\dfrac{3 - \sqrt{5}}{2} \approx -0.876$
The solution set is $x =$ {$-0.876, 0.876 $}
