College Algebra, Chapter 5, Review Exercise, Section Review Exercise, Problem 66

Evaluate the equaiton $\log x + \log (x + 1) = \log 12 $. Find the exact solution, otherwise use a calculator.

$
\begin{equation}
\begin{aligned}
\log x + \log (x + 1) &= \log 12 \\
\\
\log x ( x + 1 ) &= \log 12 && \text{Laws of Logarithm }\log_a AB = \log_a A + \log_a B\\
\\
e^{\log x(x+1)} &= e^{\log 12} && \text{Raise $e$ to each side}\\
\\
x(x+1) &= 12 && \text{Properties of log}\\
\\
x^2 + x &= 12 && \text{Distributive property}\\
\\
x^2 + x - 12 & = 0 && \text{Subtract 12 }\\
\\
(x + 4)(x - 3) &= 0 && \text{Factor}
\end{aligned}
\end{equation}
$

Solve for $x$

$
\begin{equation}
\begin{aligned}
x + 4 &= 0 &\text{and}&& x -3 &= 0\\
\\
x &= -4 &&& x &= 3
\end{aligned}
\end{equation}
$

The only solution in the given equation is $x = 3$, since $x = -4$ will give a negative value.