College Algebra, Chapter 5, 5.4, Section 5.4, Problem 54

Solve the Logarithmic Equation $\ln(x -1) + \ln (x +2) = 1$ for $x$.


$
\begin{equation}
\begin{aligned}
\ln(x -1) + \ln (x +2) &= 1\\
\\
\ln(x -1)(x+2) &= 1 && \text{Laws of Logarithms}\\
\\
e^{\ln(x-1)(x+2)} &= e^1 && \text{Raise $e$ to each side}\\
\\
(x - 1) (x + 2) &= e && \text{Expand}\\
\\
x^2 + x - 2 &= e\\
\\
x^2 + x &= e + 2 && \text{Add 2}\\
\\
x^2 + x + \frac{1}{4} &= e + 2 + \frac{1}{4} && \text{Complete the square: Add } \left( \frac{1}{2} \right)^2 = \frac{1}{4}\\
\\
\left( x + \frac{1}{2} \right)^2 &= e + \frac{9}{4} && \text{Simplify}\\
\\
x + \frac{1}{2} &= \pm \sqrt{e + \frac{9}{4}} && \text{Take the square root}\\
\\
x &= -\frac{1}{2} \pm \sqrt{e + \frac{9}{4}} && \text{Subtract } \frac{1}{2}
\end{aligned}
\end{equation}
$


We have, $\displaystyle x = -\frac{1}{2}+\sqrt{e+\frac{9}{4}} \text{ and } x = -\frac{1}{2} - \sqrt{e + \frac{9}{4}}$
The only solution of the given equation is $\displaystyle x = -\frac{1}{2} + \sqrt{e + \frac{9}{4}}$,
since the natural logarithm of a negative value is undefined.