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 \( v \).

Name: How do I solve the problem in the picture I attatched? ### Problem Sta
Uploaded: 2024-03-05T11:52:21.000Z
Channel: Bartleby
Description: 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 exp