Use logarithms to solve. $$ e^{2 x}-e^{x}-132=0 $$ Enter the exact answer (i.e. keep your answer in exponential or logarithmic form, you do not need to calculate its numeric value). Enclose arguments of functions in parentheses and include a multiplication sign between terms. For example, $$ c^{*} \ln (h) $$. Hints: - Remember by the rule of exponents, that $$ e^{2 x}=\left(e^{x}\right)^{2} $$ so that you can try to factor this as ( $$ e^{x}+ $$ something $$ )\left(e^{x}+\right. $$ something else $$ ) $$ where the "something" and "something else" can be positive or negative numbers. - Alternately, you could let $$ y=e^{x} $$ and this equation would be $$ y^{2}-y-132=0 $$ and you can solve first for $$ y $$ and then for $$ x $$. $$ x=\square \square^{\circ} $$

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Solve the equation $$ e^{2x}-e^{x}-132=0 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$e^{2x}-e^{x}-132=0$$
- step1: Factor the expression:
  $$\left(e^{x}-12ight)\left(e^{x}+11ight)=0$$
- step2: Separate into possible cases:
  $$\begin{align}&e^{x}-12=0\&e^{x}+11=0\end{align}$$
- step3: Solve the equation:
  $$\begin{align}&x=\ln{\left(12ight)}\&x 
otin \mathbb{R}\end{align}$$
- step4: Find the union:
  $$x=\ln{\left(12ight)}$$
The solution to the equation $$e^{2x}-e^{x}-132=0$$ is $$x=\ln{(12)}$$.
$$ x = \ln(12) $$

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