Answer
The solution set is $\{\log_{5}4\}$ or $\{0.861\}$
Work Step by Step
... $\quad 25x=(5^{2})^{x}=(5^{x})^{2}$
... Substitute $t=5^{x}$ (t is positive)
... the equation becomes
$t^{2}-8t+16=0 \quad $...recognize a square of a difference
$(t-4)^{2}=0$
$t=4 \quad $... bring back x
$5^{x}=4 \quad $...Apply$: \quad \log_{5}(...)$
$x=\log_{5}4\approx 0.861$
The solution set is $\{\log_{5}4\}$ or $\{0.861\}$
