Answer
The required value is $ x=2$
Work Step by Step
Consider the given equation,
$\ln \left( x+4 \right)-\ln \left( x+1 \right)=\ln x $
Apply the quotient rule of logarithms
$\begin{align}
& \ln \left( x+4 \right)-\ln \left( x+1 \right)=\ln x \\
& \ln \frac{\left( x+4 \right)}{\left( x+1 \right)}=\ln x
\end{align}$
Comparing both sides, we get,
$\begin{align}
& \frac{\left( x+4 \right)}{\left( x+1 \right)}=x \\
& \left( x+4 \right)=x\left( x+1 \right) \\
& \left( x+4 \right)={{x}^{2}}+x \\
\end{align}$
Add to both sides $-x $
$\begin{align}
& \left( x+4-x \right)={{x}^{2}}+x-x \\
& {{x}^{2}}=4
\end{align}$
Taking the square root on both sides, we get:
$\begin{align}
& \sqrt{{{x}^{2}}}=\sqrt{4} \\
& x=2,-2
\end{align}$
Here $ x=-2$ is not the solution of the given expression because
$\begin{align}
& \ln \left( x+4 \right)-\ln \left( x+1 \right)=\ln x \\
& \ln \left( -2+4 \right)-\ln \left( -2+1 \right)=\ln \left( -2 \right) \\
& \ln \left( 2 \right)-\ln \left( -1 \right)=\ln \left( -2 \right)
\end{align}$
So, by definition of ${{\log }_{a}}b $, $ b>0$ but $-1$ and $-2$ is not greater than zero
Thus, $ x=-2$ does not satisfy the equation $\ln \left( x+4 \right)-\ln \left( x+1 \right)=\ln x $
Hence, $ x=2$ is the only solution of the given expression.
Therefore, the value of $\ln \left( x+4 \right)-\ln \left( x+1 \right)=\ln x $ is $ x=2$.
