Answer
$x=e^2$
Work Step by Step
Using the properties of logarithms, the given equation, $
\ln x^2+1=5
,$ is equivalent to
\begin{align*}\require{cancel}
\ln x^2+1-1&=5-1
\\
\ln x^2&=4
\\
2\ln x&=4
&\left(\text{use } \ln a^x=x\ln a \right)
\\\\
\dfrac{2\ln x}{2}&=\dfrac{4}{2}
\\\\
\ln x&=2
\\
\log_e x&=2
&\left(\text{use } \ln x=\log_ex \right)
\\
x&=e^2
&\left(\text{use } y=b^x\Rightarrow\log_by=x \right)
.\end{align*}
Hence, the solution is $
x=e^2
.$
CHECKING: Substituting $
x=e^2
$ in the given equation, $
\ln x^2+1=5
,$ results to
\begin{align*}
\ln (e^2)^2+1&=5
\\
\ln e^4+1=&5
&\left(\text{use } (a^m)^n=a^{mn} \right)
\\
4\ln e+1=&5
&\left(\text{use } \ln a^x=x\ln a \right)
\\
4(1)+1=&5
&\left(\text{use } \ln e=1 \right)
\\
4+1=&5
\\
5=&5
&\text{ (TRUE)}
.\end{align*}
Since the substitution above ended with a TRUE statement, then $
x=e^2
$ is the solution.