Question
Question 19 Find the solution of the exponential equation \( e^{2 x+1}=37 \) in terms of logarithms, or correct to four decimal places. \( x=\square \)
Ask by Savage Campos. in the United States
Mar 13,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
\( x \approx 1.3055 \)
Solution
Solve the equation \( e^{2x+1}=37 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(e^{2x+1}=37\)
- step1: Take the logarithm of both sides:
\(\ln{\left(e^{2x+1}\right)}=\ln{\left(37\right)}\)
- step2: Evaluate the logarithm:
\(2x+1=\ln{\left(37\right)}\)
- step3: Move the constant to the right side:
\(2x=\ln{\left(37\right)}-1\)
- step4: Divide both sides:
\(\frac{2x}{2}=\frac{\ln{\left(37\right)}-1}{2}\)
- step5: Divide the numbers:
\(x=\frac{\ln{\left(37\right)}-1}{2}\)
The solution to the exponential equation \(e^{2x+1}=37\) in terms of logarithms is \(x=\frac{\ln(37)-1}{2}\). Correct to four decimal places, the solution is \(x\approx 1.3055\).
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Simplify this solution Bonus Knowledge
To solve the equation \( e^{2x+1} = 37 \), we first take the natural logarithm of both sides: \[ \ln(e^{2x+1}) = \ln(37) \] This simplifies to: \[ 2x + 1 = \ln(37) \] Next, we isolate \( x \): \[ 2x = \ln(37) - 1 \] \[ x = \frac{\ln(37) - 1}{2} \] Calculating this gives: \[ x \approx \frac{3.6109 - 1}{2} \approx \frac{2.6109}{2} \approx 1.3054 \] Thus, the solution is \( x \approx 1.3054 \) (to four decimal places).
