Question
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 \( x \). \[ x=\square \square^{\circ} \]
Ask by Moreno Ellis. in the United States
Jan 25,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
\( x = \ln(12) \)
Solution
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:
\(\left(e^{x}-12\right)\left(e^{x}+11\right)=0\)
- step2: Separate into possible cases:
\(\begin{align}&e^{x}-12=0\\&e^{x}+11=0\end{align}\)
- step3: Solve the equation:
\(\begin{align}&x=\ln{\left(12\right)}\\&x \notin \mathbb{R}\end{align}\)
- step4: Find the union:
\(x=\ln{\left(12\right)}\)
The solution to the equation \(e^{2x}-e^{x}-132=0\) is \(x=\ln{(12)}\).
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Simplify this solution The Deep Dive
To solve the equation \( e^{2x} - e^{x} - 132 = 0 \), we can make the substitution \( y = e^{x} \), transforming our equation into \( y^{2} - y - 132 = 0 \). This is a quadratic equation that can be factored as \( (y - 12)(y + 11) = 0 \). Therefore, we have the solutions \( y = 12 \) and \( y = -11 \). Since \( e^{x} \) cannot be negative, we only consider \( y = 12 \). Thus, \( e^{x} = 12 \). Applying the natural logarithm, we find \( x = \ln (12) \). \[ x = \ln (12) \]
