Chapter 5 - Exponential and Logarithmic Functions - 5.4 Logarithmic Functions - 5.4 Assess Your Understanding - Page 296: 105

$\pm2\sqrt{2}$

We know that by definition then $log_a {y}=x$, then $y=a^x$. Hence if $log_3 {(x^2+1)}=2$, then $3^{2}=9=x^2+1$. Solve the equation above to obtain: \begin{align*}8=x^2\end{align*} \begin{align*}x=\pm\sqrt8=\pm2\sqrt{2}\end{align*}