How do you solve $\ln \left( -3x-1 \right)-\ln 7=2$?

Hint: We solve the given equation using the different identity formulas of logarithm like $\ln a-\ln b=\ln \dfrac{a}{b}$, ${{\log }_{e}}a=y\Rightarrow a={{e}^{y}}$. The main step would be to eliminate the logarithm function and keep only the linear equation of x. We solve the linear equation with the help of basic binary operations.

Complete step by step answer:
We take the logarithmic identity for the given equation $\ln \left( -3x-1 \right)-\ln 7=2$ to find the solution for x.
We have $\ln a={{\log }_{e}}a$. The subtraction for logarithm works as $\ln a-\ln b=\ln \dfrac{a}{b}$.
We operate the subtraction part in the left-hand side of $\ln \left( -3x-1 \right)-\ln 7=2$.
$\begin{align}
  & \ln \left( -3x-1 \right)-\ln 7 \\
 & =\ln \dfrac{\left( -3x-1 \right)}{7} \\
\end{align}$
We get a single equation of logarithm and a constant on the right-hand side
Therefore, $\ln \dfrac{\left( -3x-1 \right)}{7}={{\log }_{e}}\dfrac{\left( -3x-1 \right)}{7}=2$.
Now we have to eliminate the logarithm function to find the value of x.
We know ${{\log }_{e}}a=y\Rightarrow a={{e}^{y}}$. Applying the rule in case of ${{\log }_{e}}\dfrac{\left( -3x-1 \right)}{7}=2$, we get
$\begin{align}
  & {{\log }_{e}}\dfrac{\left( -3x-1 \right)}{7}=2 \\
 & \Rightarrow \dfrac{\left( -3x-1 \right)}{7}={{e}^{2}} \\
\end{align}$
Now we have a linear equation of x. We need to solve it.
We multiply with -7 both side of the equation $\dfrac{\left( -3x-1 \right)}{7}={{e}^{2}}$ and get $-7\left[ \dfrac{\left( -3x-1 \right)}{7} \right]=-7{{e}^{2}}$.
$\begin{align}
  & -7\left[ \dfrac{\left( -3x-1 \right)}{7} \right]=-7{{e}^{2}} \\
 & \Rightarrow 3x+1=-7{{e}^{2}} \\
\end{align}$
We solve the value of x as
$\begin{align}
  & 3x+1=-7{{e}^{2}} \\
 & \Rightarrow 3x=-7{{e}^{2}}-1 \\
 & \Rightarrow x=\dfrac{-7{{e}^{2}}-1}{3} \\
\end{align}$

Therefore, the solution for the equation $\ln \left( -3x-1 \right)-\ln 7=2$ is $x=\dfrac{-7{{e}^{2}}-1}{3}$.

Note: In case of the base is not mentioned then the general solution for the base for logarithm is 10. But the base of $e$ is fixed for $\ln $.
We also need to remember that for logarithm function there has to be a domain constraint.
For any ${{\log }_{e}}a$, $a>0$. This means for $\ln \left( -3x-1 \right)$, $\left( -3x-1 \right)>0$.
The simplified form is $x<-\dfrac{1}{3}$.