Answer
The required solution is $x=3$
Work Step by Step
$\log x+\log (x+1)=\log 12$
By using the product rule of the logarithm:
$\log a+\log b=\log \left( ab \right)$ ,
$\begin{align}
& \log (x(x+1))=\log 12 \\
& \log (x.x+x.1)=\log 12
\end{align}$
Also, using the power rule,
$\begin{align}
& \log ({{x}^{1+1}}+x.1)=\log 12 \\
& \log ({{x}^{2}}+x)=\log 12
\end{align}$
Therefore, from the above equation, the argument of the logarithm on both side of the equation must be equal.
$\begin{align}
& ({{x}^{2}}+x)=12 \\
& {{x}^{2}}+x-12=0 \\
& (x-3)(x+4)=0
\end{align}$
Then, equating the equations,
$\begin{align}
& (x-3)=0 \\
& x=3 \\
& (x+4)=0 \\
& x=-4
\end{align}$
Thus, by putting the values of x and verifying the original equation, $x=-4$ was proven to be invalid.
