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
![**Problem Statement:**
Let \( \log_b 7 = Z \) and \( \log_b 4 = L \). Write the expression in terms of \( Z \) and \( L \).
**Expression:**
\[ \log_b \sqrt{\frac{7}{64}} \]
**Solution Setup:**
\[ \log_b \sqrt{\frac{7}{64}} = \Box \]
---
**Explanation:**
To solve for the box, note that:
1. The expression inside the logarithm can be rewritten using the properties of logarithms and exponents.
2. \( \sqrt{\frac{7}{64}} = \left(\frac{7}{64}\right)^{1/2} = 7^{1/2} \times 64^{-1/2} \).
3. Therefore, \( \log_b \left( 7^{1/2} \times 64^{-1/2} \right) \) becomes \( \frac{1}{2} \log_b 7 - \frac{1}{2} \log_b 64 \).
Using the given \( \log_b 7 = Z \) and evaluating \( \log_b 64 \), knowing \( 64 = 4^3 \):
\[ \log_b 64 = \log_b 4^3 = 3\log_b 4 = 3L \]
Now substitute:
\[ \frac{1}{2}(Z) - \frac{1}{2}(3L) \]
\[ \frac{1}{2}Z - \frac{3}{2}L \]
Therefore, the answer is:
\[ \boxed{\frac{1}{2}Z - \frac{3}{2}L} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fee5144b1-2276-4010-857d-2387de3c8638%2F92fafe11-ab64-4cc3-8bed-e10ae17d2335%2Fxxlxsj2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Let \( \log_b 7 = Z \) and \( \log_b 4 = L \). Write the expression in terms of \( Z \) and \( L \).
**Expression:**
\[ \log_b \sqrt{\frac{7}{64}} \]
**Solution Setup:**
\[ \log_b \sqrt{\frac{7}{64}} = \Box \]
---
**Explanation:**
To solve for the box, note that:
1. The expression inside the logarithm can be rewritten using the properties of logarithms and exponents.
2. \( \sqrt{\frac{7}{64}} = \left(\frac{7}{64}\right)^{1/2} = 7^{1/2} \times 64^{-1/2} \).
3. Therefore, \( \log_b \left( 7^{1/2} \times 64^{-1/2} \right) \) becomes \( \frac{1}{2} \log_b 7 - \frac{1}{2} \log_b 64 \).
Using the given \( \log_b 7 = Z \) and evaluating \( \log_b 64 \), knowing \( 64 = 4^3 \):
\[ \log_b 64 = \log_b 4^3 = 3\log_b 4 = 3L \]
Now substitute:
\[ \frac{1}{2}(Z) - \frac{1}{2}(3L) \]
\[ \frac{1}{2}Z - \frac{3}{2}L \]
Therefore, the answer is:
\[ \boxed{\frac{1}{2}Z - \frac{3}{2}L} \]
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