Answer
$x=4$
Work Step by Step
Consider the given expression:
$\log \left( x+4 \right)-\log \left( x-2 \right)=\log x$
Using the product identity, we get
${{\log }_{a}}A+{{\log }_{b}}B={{\log }_{ab}}(AB)$
So,
$\begin{align}
& \log \left( x+4 \right)-\log (x-2)=\log x \\
& \log \left( x+4 \right)=\log x+\log \left( x-2 \right) \\
& \log \left( x+4 \right)=\log \left( x\left( x-2 \right) \right)
\end{align}$
Use anti log.
Then,
$\begin{align}
& \left( x+4 \right)={{x}^{2}}-2x \\
& {{x}^{2}}-2x=x+4 \\
& {{x}^{2}}-2x-x-4=0 \\
& {{x}^{2}}-3x-4=0
\end{align}$
This equation can be solved by the quadratic equation of the form $a{{x}^{2}}+bx+c=0$
$\begin{align}
& \frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
& a=1,b=-3,c=-4 \\
& {{x}_{1,2}}=\frac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\times 1\times \left( -4 \right)}}{2\times 1} \\
& x=4,x=-1
\end{align}$
We eliminate the $-1$ solution because we can not take the log of a negative number.
