. Let u = Inx and v = Iny.) write the given expression in terms of u and v. In(Vx •y) (hint, first expend the In, then plug in u and v.

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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How do I solve the problem in the picture I attatched?

### Problem Statement:

1. Let \( u = \ln(x) \) and \( v = \ln(y) \). Write the given expression in terms of \( u \) and \( v \).

   \[
   \ln\left(\sqrt{x} \cdot y^2\right) 
   \]

   *(Hint: first expand the \(\ln\), then plug in \( u \) and \( v \)).*

### Solution:

To solve the problem, follow these steps:

1. **Recognize the expression that needs to be expanded:**

   \[
   \ln\left(\sqrt{x} \cdot y^2\right)
   \]

2. **Use logarithmic properties to expand the expression:**

   - The property \(\ln(a \cdot b) = \ln(a) + \ln(b)\) lets us break down the product:
     \[
     \ln\left(\sqrt{x} \cdot y^2\right) = \ln(\sqrt{x}) + \ln(y^2)
     \]

   - The property \(\ln(a^b) = b \cdot \ln(a)\) helps with exponents:
     \[
     \ln(\sqrt{x}) = \ln(x^{1/2}) = \frac{1}{2} \ln(x)
     \]
     \[
     \ln(y^2) = 2 \ln(y)
     \]

3. **Substitute the expressions for \( \ln(x) \) and \( \ln(y) \):**

   - Substitute \( u = \ln(x) \) and \( v = \ln(y) \) into the expanded expression:
     \[
     \frac{1}{2} \ln(x) + 2 \ln(y) = \frac{1}{2} u + 2v
     \]

4. **Final expression in terms of \( u \) and \( v \):**

   \[
   \frac{1}{2} u + 2v
   \]

This expression is the solution written in terms of the given variables \( u \) and \( v \).
Transcribed Image Text:### Problem Statement: 1. Let \( u = \ln(x) \) and \( v = \ln(y) \). Write the given expression in terms of \( u \) and \( v \). \[ \ln\left(\sqrt{x} \cdot y^2\right) \] *(Hint: first expand the \(\ln\), then plug in \( u \) and \( v \)).* ### Solution: To solve the problem, follow these steps: 1. **Recognize the expression that needs to be expanded:** \[ \ln\left(\sqrt{x} \cdot y^2\right) \] 2. **Use logarithmic properties to expand the expression:** - The property \(\ln(a \cdot b) = \ln(a) + \ln(b)\) lets us break down the product: \[ \ln\left(\sqrt{x} \cdot y^2\right) = \ln(\sqrt{x}) + \ln(y^2) \] - The property \(\ln(a^b) = b \cdot \ln(a)\) helps with exponents: \[ \ln(\sqrt{x}) = \ln(x^{1/2}) = \frac{1}{2} \ln(x) \] \[ \ln(y^2) = 2 \ln(y) \] 3. **Substitute the expressions for \( \ln(x) \) and \( \ln(y) \):** - Substitute \( u = \ln(x) \) and \( v = \ln(y) \) into the expanded expression: \[ \frac{1}{2} \ln(x) + 2 \ln(y) = \frac{1}{2} u + 2v \] 4. **Final expression in terms of \( u \) and \( v \):** \[ \frac{1}{2} u + 2v \] This expression is the solution written in terms of the given variables \( u \) and \( v \).
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