How do you solve $\ln {{\left( x+1 \right)}^{2}}=2$?

Hint: In the problem we have the function $\ln $ which is nothing but a logarithmic function with base as $e$. So, we will write the $\ln $ function as the logarithmic function. Now we will use the exponential laws and convert the given equation in terms of exponential. Now we will apply the square root on both sides of the obtained function and simplify then we will get the result.

Complete step-by-step answer:
Given that, $\ln {{\left( x+1 \right)}^{2}}=2$.
We know that $\ln X={{\log }_{e}}X$, using this formula in the given equation, then we will get
$\begin{align}
  & \ln {{\left( x+1 \right)}^{2}}=2 \\
 & \Rightarrow {{\log }_{e}}{{\left( x+1 \right)}^{2}}=2 \\
\end{align}$
We have the basic logarithmic rule i.e., ${{\log }_{e}}X=Y\Leftrightarrow X={{e}^{Y}}$. Using this formula in the above equation, then we will get
$\begin{align}
  & {{\log }_{e}}{{\left( x+1 \right)}^{2}}=2 \\
 & \Rightarrow {{\left( x+1 \right)}^{2}}={{e}^{2}} \\
\end{align}$
Applying square root on both sides of the above equation, then we will have
$\sqrt{{{\left( x+1 \right)}^{2}}}=\sqrt{{{e}^{2}}}$
We know that the square and square root are the inverse function and they will be cancelled when they are in multiplication. Hence, we will get
$x+1=e$
Subtracting one from the both sides of the above equation, then we will get
$x+1-1=e-1$
We know that $+X-X=0$, then we will get
$\therefore x=e-1$

Note: We can also solve the problem in another method. In this method we will use the logarithmic formula $\ln {{x}^{a}}=a\ln x$ and simplify the obtained equation by using the basic logarithmic formulas.
Given that, $\ln {{\left( x+1 \right)}^{2}}=2$
Applying $\ln {{x}^{a}}=a\ln x$ in the above equation, then we will get
$2\ln \left( x+1 \right)=2$
Dividing the above equation with $2$ on both sides.
$\ln \left( x+1 \right)=1$
We have the basic logarithmic formula $\ln X=Y\Leftrightarrow X={{e}^{Y}}$, then we will get
$\begin{align}
  & \Rightarrow x+1={{e}^{1}} \\
 & \Rightarrow x+1=e \\
 & \Rightarrow x=e-1 \\
\end{align}$
From both the methods we got the same result.