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.