If `"log"_(4) 5 = x " and log"_(5) 6 = y, " then log"_(2) 3` is equal to

Options

A. 2xy + 1
B. 2xy-1
C. 2x+1
D. 2y + 1

Explanation

To solve the problem, we need to find the value of \( \log_2 3 \) in terms of \( x \) and \( y \), where \( x = \log_4 5 \) and \( y = \log_5 6 \).

### Step-by-Step Solution:

1. **Express \( x \) and \( y \) using the change of base formula:**
\[ x = \log_4 5 = \frac{\log 5}{\log 4} \]
\[ y = \log_5 6 = \frac{\log 6}{\log 5} \]

2. **Multiply \( x \) and \( y \):**
\[ xy = \left(\frac{\log 5}{\log 4}\right) \left(\frac{\log 6}{\log 5}\right) = \frac{\log 6}{\log 4} \]

3. **Rewrite \( \log 6 \) in terms of \( \log 3 \) and \( \log 2 \):**
\[ \log 6 = \log(3 \cdot 2) = \log 3 + \log 2 \]
Therefore,
\[ xy = \frac{\log 3 + \log 2}{\log 4} \]

4. **Express \( \log 4 \) in terms of \( \log 2 \):**
\[ \log 4 = \log(2^2) = 2 \log 2 \]
Thus,
\[ xy = \frac{\log 3 + \log 2}{2 \log 2} \]

5. **Rearranging gives:**
\[ 2xy \log 2 = \log 3 + \log 2 \]

6. **Isolate \( \log 3 \):**
\[ \log 3 = 2xy \log 2 - \log 2 \]
\[ \log 3 = (2xy - 1) \log 2 \]

7. **Divide both sides by \( \log 2 \) to find \( \log_2 3 \):**
\[ \log_2 3 = 2xy - 1 \]

### Final Answer:
\[ \log_2 3 = 2xy - 1 \]