Answer
$x=2$
Work Step by Step
Consider the given expression:
${{\log }_{2}}x+{{\log }_{2}}\left( x+2 \right)=3$
Using the product identity, we get
${{\log }_{a}}A+{{\log }_{b}}B={{\log }_{ab}}\left( AB \right)$
So,
$\begin{align}
& {{\log }_{2}}x+{{\log }_{2}}\left( x+2 \right)=3 \\
& \log \left( x\left( x+2 \right) \right)=3
\end{align}$
Use anti log.
Then,
$\begin{align}
& {{\log }_{2}}\left( x\left( x+2 \right) \right)=3 \\
& x\left( x+2 \right)={{2}^{3}} \\
& {{x}^{2}}+2x=8 \\
& {{x}^{2}}+2x-8=0
\end{align}$
This implies that.
$\begin{align}
& {{x}^{2}}-2x+4x-8=0 \\
& x(x-2)+4(x-2)=0 \\
& \left( x-2 \right)\left( x+4 \right)=0 \\
& x=2,x=-4
\end{align}$
We eliminate the $-4$ solution, because we can not take the log of a negative number.
