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,...
Related questions
Question
Please help with both equations thank you
![**Problem Statement**
Solve each equation. Round to two decimal places as needed. Show all of your work.
1. \(5 \ln(3x) + 1 = 16\)
2. \(5(10^{3x}) + 1 = 16\)
---
**Equation 1: Solving \(5 \ln(3x) + 1 = 16\)**
1. **Subtract 1 from both sides:**
\[
5 \ln(3x) = 15
\]
2. **Divide both sides by 5:**
\[
\ln(3x) = 3
\]
3. **Exponentiate both sides to solve for \(3x\):**
\[
3x = e^3
\]
4. **Calculate \(e^3\) and solve for \(x\):**
\[
3x = 20.09 \quad (approximately)
\]
5. **Divide by 3:**
\[
x = \frac{20.09}{3} \approx 6.70
\]
**Equation 2: Solving \(5(10^{3x}) + 1 = 16\)**
1. **Subtract 1 from both sides:**
\[
5(10^{3x}) = 15
\]
2. **Divide both sides by 5:**
\[
10^{3x} = 3
\]
3. **Take the logarithm base 10 of both sides:**
\[
3x \log_{10}(10) = \log_{10}(3)
\]
Simplifying, since \(\log_{10}(10) = 1\):
\[
3x = \log_{10}(3)
\]
4. **Calculate \(\log_{10}(3)\) and solve for \(x\):**
\[
3x = 0.4771 \quad (approximately)
\]
5. **Divide by 3:**
\[
x = \frac{0.4771}{3} \approx 0.16
\]
---
**Conclusion**
The solutions are:
- For the first equation, \(x \approx](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbd0d1a38-c045-41ea-8778-69e62db1599e%2Fcb0d3cf8-4a9a-4992-bc12-7d4be1b53648%2F5j8imgr_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement**
Solve each equation. Round to two decimal places as needed. Show all of your work.
1. \(5 \ln(3x) + 1 = 16\)
2. \(5(10^{3x}) + 1 = 16\)
---
**Equation 1: Solving \(5 \ln(3x) + 1 = 16\)**
1. **Subtract 1 from both sides:**
\[
5 \ln(3x) = 15
\]
2. **Divide both sides by 5:**
\[
\ln(3x) = 3
\]
3. **Exponentiate both sides to solve for \(3x\):**
\[
3x = e^3
\]
4. **Calculate \(e^3\) and solve for \(x\):**
\[
3x = 20.09 \quad (approximately)
\]
5. **Divide by 3:**
\[
x = \frac{20.09}{3} \approx 6.70
\]
**Equation 2: Solving \(5(10^{3x}) + 1 = 16\)**
1. **Subtract 1 from both sides:**
\[
5(10^{3x}) = 15
\]
2. **Divide both sides by 5:**
\[
10^{3x} = 3
\]
3. **Take the logarithm base 10 of both sides:**
\[
3x \log_{10}(10) = \log_{10}(3)
\]
Simplifying, since \(\log_{10}(10) = 1\):
\[
3x = \log_{10}(3)
\]
4. **Calculate \(\log_{10}(3)\) and solve for \(x\):**
\[
3x = 0.4771 \quad (approximately)
\]
5. **Divide by 3:**
\[
x = \frac{0.4771}{3} \approx 0.16
\]
---
**Conclusion**
The solutions are:
- For the first equation, \(x \approx
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