**Solving Exponential Equations Using Logarithms** To solve the given exponential equation, follow these steps: \[ e^{2x + 1} = 40 \] We want to solve for \( x \) in terms of logarithms or correct to four decimal places. 1. **Isolate the exponential term:** \[ e^{2x + 1} = 40 \] 2. **Take the natural logarithm on both sides:** \[ \ln(e^{2x + 1}) = \ln(40) \] 3. **Simplify using the property of logarithms:** \[ 2x + 1 = \ln(40) \] 4. **Solve for \( x \):** \[ 2x + 1 = \ln(40) \] \[ 2x = \ln(40) - 1 \] \[ x = \frac{\ln(40) - 1}{2} \] Thus, the solution in terms of logarithms is: \[ x = \frac{\ln(40) - 1}{2} \] To find the numerical value correct to four decimal places, use a calculator to evaluate \( \ln(40) \): \[ \ln(40) \approx 3.6889 \] \[ x \approx \frac{3.6889 - 1}{2} \approx \frac{2.6889}{2} \approx 1.3445 \] Therefore, the solution to four decimal places is: \[ x \approx 1.3445 \] Click the "**Next Question**" button to proceed to the next problem.
**Solving Exponential Equations Using Logarithms** To solve the given exponential equation, follow these steps: \[ e^{2x + 1} = 40 \] We want to solve for \( x \) in terms of logarithms or correct to four decimal places. 1. **Isolate the exponential term:** \[ e^{2x + 1} = 40 \] 2. **Take the natural logarithm on both sides:** \[ \ln(e^{2x + 1}) = \ln(40) \] 3. **Simplify using the property of logarithms:** \[ 2x + 1 = \ln(40) \] 4. **Solve for \( x \):** \[ 2x + 1 = \ln(40) \] \[ 2x = \ln(40) - 1 \] \[ x = \frac{\ln(40) - 1}{2} \] Thus, the solution in terms of logarithms is: \[ x = \frac{\ln(40) - 1}{2} \] To find the numerical value correct to four decimal places, use a calculator to evaluate \( \ln(40) \): \[ \ln(40) \approx 3.6889 \] \[ x \approx \frac{3.6889 - 1}{2} \approx \frac{2.6889}{2} \approx 1.3445 \] Therefore, the solution to four decimal places is: \[ x \approx 1.3445 \] Click the "**Next Question**" button to proceed to the next problem.
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Question
![**Solving Exponential Equations Using Logarithms**
To solve the given exponential equation, follow these steps:
\[ e^{2x + 1} = 40 \]
We want to solve for \( x \) in terms of logarithms or correct to four decimal places.
1. **Isolate the exponential term:**
\[ e^{2x + 1} = 40 \]
2. **Take the natural logarithm on both sides:**
\[ \ln(e^{2x + 1}) = \ln(40) \]
3. **Simplify using the property of logarithms:**
\[ 2x + 1 = \ln(40) \]
4. **Solve for \( x \):**
\[ 2x + 1 = \ln(40) \]
\[ 2x = \ln(40) - 1 \]
\[ x = \frac{\ln(40) - 1}{2} \]
Thus, the solution in terms of logarithms is:
\[ x = \frac{\ln(40) - 1}{2} \]
To find the numerical value correct to four decimal places, use a calculator to evaluate \( \ln(40) \):
\[ \ln(40) \approx 3.6889 \]
\[ x \approx \frac{3.6889 - 1}{2} \approx \frac{2.6889}{2} \approx 1.3445 \]
Therefore, the solution to four decimal places is:
\[ x \approx 1.3445 \]
Click the "**Next Question**" button to proceed to the next problem.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe2c5c293-7ff8-490b-b548-13e2f43ca7a8%2Facd6b863-17b4-4ed4-b393-612990ef54a1%2F664ba3t.jpeg&w=3840&q=75)
Transcribed Image Text:**Solving Exponential Equations Using Logarithms**
To solve the given exponential equation, follow these steps:
\[ e^{2x + 1} = 40 \]
We want to solve for \( x \) in terms of logarithms or correct to four decimal places.
1. **Isolate the exponential term:**
\[ e^{2x + 1} = 40 \]
2. **Take the natural logarithm on both sides:**
\[ \ln(e^{2x + 1}) = \ln(40) \]
3. **Simplify using the property of logarithms:**
\[ 2x + 1 = \ln(40) \]
4. **Solve for \( x \):**
\[ 2x + 1 = \ln(40) \]
\[ 2x = \ln(40) - 1 \]
\[ x = \frac{\ln(40) - 1}{2} \]
Thus, the solution in terms of logarithms is:
\[ x = \frac{\ln(40) - 1}{2} \]
To find the numerical value correct to four decimal places, use a calculator to evaluate \( \ln(40) \):
\[ \ln(40) \approx 3.6889 \]
\[ x \approx \frac{3.6889 - 1}{2} \approx \frac{2.6889}{2} \approx 1.3445 \]
Therefore, the solution to four decimal places is:
\[ x \approx 1.3445 \]
Click the "**Next Question**" button to proceed to the next problem.
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