If `log_(4)5 = x` and `log_(5)6 = y` then

Options

A. `log_(4)6 = xy`
B. `log_(6)4 = xy`
C. `log_(3)2 =(1)/(2xx y-1)`
D. `log_(2)3 (1)/(2x y -1)`

Answer

To solve the problem, we need to find the values of \( xy \) and \( \frac{1}{2xy - 1} \) given that \( \log_{4}5 = x \) and \( \log_{5}6 = y \).

### Step 1: Find \( xy \)

We start with the expressions for \( x \) and \( y \):
\[ x = \log_{4}5 \quad \text{and} \quad y = \log_{5}6 \]

Using the property of logarithms that states \( \log_{a}b = \frac{\log_{c}b}{\log_{c}a} \), we can express \( x \) and \( y \) in terms of a common base, say base \( \alpha \):
\[ x = \frac{\log_{\alpha}5}{\log_{\alpha}4} \quad \text{and} \quad y = \frac{\log_{\alpha}6}{\log_{\alpha}5} \]

Now, we can find \( xy \):
\[ xy = \left(\frac{\log_{\alpha}5}{\log_{\alpha}4}\right) \left(\frac{\log_{\alpha}6}{\log_{\alpha}5}\right) \]

The \( \log_{\alpha}5 \) terms cancel out:
\[ xy = \frac{\log_{\alpha}6}{\log_{\alpha}4} \]

Using the change of base formula again, we can rewrite this as:
\[ xy = \log_{4}6 \]

### Step 2: Find \( \frac{1}{2xy - 1} \)

Now we need to calculate \( \frac{1}{2xy - 1} \):
\[ \frac{1}{2xy - 1} = \frac{1}{2\log_{4}6 - 1} \]

We can express \( 2\log_{4}6 \) as:
\[ 2\log_{4}6 = \log_{4}6^2 \]

Thus, we have:
\[ \frac{1}{2xy - 1} = \frac{1}{\log_{4}36 - 1} \]

Using the property of logarithms, we can rewrite \( \log_{4}36 \) as:
\[ \log_{4}36 = \log_{4}(4 \cdot 9) = \log_{4}4 + \log_{4}9 = 1 + \log_{4}9 \]

So, we can substitute this back:
\[ \frac{1}{\log_{4}36 - 1} = \frac{1}{\log_{4}9} \]

Using the change of base formula again, we have:
\[ \frac{1}{\log_{4}9} = \log_{9}4 \]

### Final Result

Thus, the final results are:
1. \( xy = \log_{4}6 \)
2. \( \frac{1}{2xy - 1} = \log_{9}4 \)