## College Algebra (10th Edition)

The solution set is $\left\{-2\sqrt{2}, 2\sqrt{2}\right\}$.
RECALL: $\log_a{x}=y \longrightarrow a^y=x, a\gt0, a\ne1$ Use the rule above to obtain: $\log_3{(x^2+1)}=2 \\\longrightarrow 3^2=x^2+1 \\9=x^2+1$ Subtract by 1 on both sides of the equation to obtain: $9-1=x^2+1-1 \\8=x^2$ Take the square root of both sides to obtain: $\pm \sqrt{8} = x \\\pm\sqrt{4(2)}=x \\\pm \sqrt{2^2(2)} = x$ Simplify the radical to obtain: $\pm2\sqrt{2}=x$ Thus, the solution set is $\left\{-2\sqrt{2}, 2\sqrt{2}\right\}$.