Answer
$x=5$ or $x=-5$
Work Step by Step
By the principle of logarithmic equality,
$\log_{3}(x^{2}+1)=\log_{3}26\quad \Rightarrow\quad x^{2}+1=26$
... subtract 1 from both sides...
$x^{2}=25$
$x=\pm 5$
(The original equation is defined for either $x=5$ or $x=-5$, so they are valid solutions.)
