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

$x \approx 0.613$
Re-write the first term as: $36^x=(6^{2})^{x}=(6^{x})^{2}$ We wish to substitute $a=6^{x}$, so the equation becomes: $a^{2}-6a+9=0$ This gives a quadratic equation, whose factors are: $(t-3)^2=0$ Use the zero factor property to obtain: $t- 3 =0 \implies t= 3$ Therefore, $6^{x}=3$ or, $x= \log_6 3 \approx 0.613$ Thus, our answer is: $x \approx 0.613$