14²x = 8 x+1 возмрах будин (2x) ly 4 = (x+1) 8 - lg 4 (x+1) • 2x = log Y = 2x (x+1)=lgy - 2x² + 2x² 20 lgs 2x 1088 2x legs logy (x+1) ligs 177% 0672x+1 •67-1=x41-1
14²x = 8 x+1 возмрах будин (2x) ly 4 = (x+1) 8 - lg 4 (x+1) • 2x = log Y = 2x (x+1)=lgy - 2x² + 2x² 20 lgs 2x 1088 2x legs logy (x+1) ligs 177% 0672x+1 •67-1=x41-1
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
How do you solve #s: 1-3.
1. How should each problem be set?
2. Then, Once each problem is set up… how do you use the log keys on the calculator to solve each problem?
![### Exponential and Logarithmic Equation Solving Example
**Problem:**
Solve the equation \(4^{2x} = 8^{x+1}\).
**Solution Steps:**
1. Start with the given equation:
\[ 4^{2x} = 8^{x+1} \]
2. Apply the logarithm to both sides to facilitate solving for \(x\):
\[ \log(4^{2x}) = \log(8^{x+1}) \]
3. Use the logarithmic power rule \(\log(a^b) = b \log(a)\) to simplify both sides:
\[ (2x) \log 4 = (x+1) \log 8 \]
4. Rewrite the equation to isolate terms involving \(x\):
\[ 2x \log 4 = (x+1) \log 8 \]
5. Distribute \(\log 8\) on the right-hand side:
\[ 2x \log 4 = x \log 8 + \log 8 \]
6. Collect all terms involving \(x\) on one side of the equation:
\[ 2x \log 4 - x \log 8 = \log 8 \]
7. Factor out the \(x\) term from the left side:
\[ x(2 \log 4 - \log 8) = \log 8 \]
8. Simplify the expression inside the parentheses using logarithmic properties:
\[ \log 4 = \log 2^2 = 2 \log 2 \]
\[ \log 8 = \log 2^3 = 3 \log 2 \]
\[ 2 (2 \log 2) - 3 \log 2 = 4 \log 2 - 3 \log 2 = \log 2 \]
9. Substitute back into the equation:
\[ x (\log 2) = \log 8 \]
10. Finally, solve for \(x\):
\[ x = \frac{\log 8}{\log 2} = 3 \]
By following these steps, we find that the solution to the equation \(4^{2x} = 8^{x + 1}\) is \(x = 3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc965ec1f-6ecd-467d-b5f1-022eb1dfa0a4%2F3a4cfc0f-7abe-4be5-aab7-468be8760246%2Fyc6vldd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Exponential and Logarithmic Equation Solving Example
**Problem:**
Solve the equation \(4^{2x} = 8^{x+1}\).
**Solution Steps:**
1. Start with the given equation:
\[ 4^{2x} = 8^{x+1} \]
2. Apply the logarithm to both sides to facilitate solving for \(x\):
\[ \log(4^{2x}) = \log(8^{x+1}) \]
3. Use the logarithmic power rule \(\log(a^b) = b \log(a)\) to simplify both sides:
\[ (2x) \log 4 = (x+1) \log 8 \]
4. Rewrite the equation to isolate terms involving \(x\):
\[ 2x \log 4 = (x+1) \log 8 \]
5. Distribute \(\log 8\) on the right-hand side:
\[ 2x \log 4 = x \log 8 + \log 8 \]
6. Collect all terms involving \(x\) on one side of the equation:
\[ 2x \log 4 - x \log 8 = \log 8 \]
7. Factor out the \(x\) term from the left side:
\[ x(2 \log 4 - \log 8) = \log 8 \]
8. Simplify the expression inside the parentheses using logarithmic properties:
\[ \log 4 = \log 2^2 = 2 \log 2 \]
\[ \log 8 = \log 2^3 = 3 \log 2 \]
\[ 2 (2 \log 2) - 3 \log 2 = 4 \log 2 - 3 \log 2 = \log 2 \]
9. Substitute back into the equation:
\[ x (\log 2) = \log 8 \]
10. Finally, solve for \(x\):
\[ x = \frac{\log 8}{\log 2} = 3 \]
By following these steps, we find that the solution to the equation \(4^{2x} = 8^{x + 1}\) is \(x = 3
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