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.