College Algebra, Chapter 5, 5.4, Section 5.4, Problem 32

Solve the equation $e^x - 12e^{-x} - 1 = 0$

$
\begin{equation}
\begin{aligned}
e^x - 12e^{-x} - 1 &= 0\\
\\
e^x - \frac{12}{e^x} - 1 &= 0 && \text{Law of exponent } a^{-n} = \frac{1}{a^n}\\
\\
\frac{e^{2x}-12-e^x}{e^x} &= 0 && \text{Get the LCD}\\
\\
e^{2x}-e^x-12 &= 0 && \text{Multiply both sides by } e^x\\
\\
(e^x - 4)(e^x + 3) &= 0 && \text{Factor using trial and error}
\end{aligned}
\end{equation}
$

Solve for $x$

$
\begin{equation}
\begin{aligned}
e^x - 4 &= 0
&&&
e^x + 3 &= 0\\
\\
e^x &= 4
&&\text{Add 4}&
e^x &= -3 && \text{Subtract 3}\\
\\
\ln e^x &= \ln 4
&&\text{Take $\ln$ of each side}&
\ln e^x &= \ln(-3) && \text{Take $\ln$ of each side}\\
\\
x &= \ln4
&&\text{Property of $\ln$}&
x &= \ln(-3) && \text{Property of $\ln$}
\end{aligned}
\end{equation}
$

The only solution to the given equation is only $x = \ln 4$, since $\ln (-3)$ is undefined.