## Precalculus (6th Edition) Blitzer

The required solution is $x=3$
$\log x+\log (x+1)=\log 12$ 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.
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