Answer
$x \approx 0.861$
Work Step by Step
Re-write the first term as:
$25^x=(5^{2})^{x}=(5^{x})^{2}$
We wish to substitute $a=5^{x}$, so the equation becomes:
$a^{2}-8a+16=0 $
This gives a quadratic equation whose factors are: $(t-4)^2=0$
Use the zero factor property to obtain: $ t-4 =0 \implies t=4$
Therefore, $5^{x}=4$
or, $x= \log_5 4 \approx 0.861$
Thus, our answer is: $x \approx 0.861$
