Question

If \(log_45=(log_4y)(log_6\sqrt5)\),then \(y\) equals

Answer 1

Given:
\[ \log_4 5 = (\log_4 y)(\log_6 \sqrt{5}) \] 

Using the change of base formula:
\[ \log_a b = \frac{\log b}{\log a} \]

\[ \log_6 \sqrt{5} = \frac{\log \sqrt{5}}{\log 6} \quad \text{and} \quad \log \sqrt{5} = \frac{1}{2} \log 5 \]

So, \[ \log_6 \sqrt{5} = \frac{1}{2} \cdot \frac{\log 5}{\log 6} \]

Now substitute into the original equation: \[ \log_4 5 = (\log_4 y) \cdot \left(\frac{1}{2} \cdot \frac{\log 5}{\log 6}\right) \]

Use change of base again for the left-hand side: \[ \log_4 5 = \frac{\log 5}{\log 4} \]

Now the equation becomes: \[ \frac{\log 5}{\log 4} = (\log_4 y) \cdot \frac{1}{2} \cdot \frac{\log 5}{\log 6} \]

Cancel \( \log 5 \) from both sides: \[ \frac{1}{\log 4} = \frac{1}{2 \log 6} \cdot \log_4 y \]

Multiply both sides by \( 2 \log 6 \): \[ \frac{2 \log 6}{\log 4} = \log_4 y \]

Now express this as a single logarithm: \[ \log_4 y = \frac{\log 4}{\log 36} \quad \text{since} \quad (2 \log 6 = \log 36) \]

Using change of base again: \[ \log_4 y = \frac{\log 4}{\log 36} = \log_{36} 4 \Rightarrow y = 36 \]

∴ The value of \( y \) is 36.

Answer 2

Given: \(\log_4 5 = (\log_4 y)(\log_6 \sqrt{5})\) 

Taking the reciprocal of \(\log_6 \sqrt{5}\) to move it to the other side: 
\(\frac{\log_4 5}{\log_6 \sqrt{5}} = \log_4 y\)

Using change of base: \(\log_6 \sqrt{5} = \frac{\log_5 \sqrt{5}}{\log_5 6} = \frac{1/2}{\log_5 6} = \frac{1}{2\log_5 6}\)

So, \(\log_4 y = \log_4 5 \cdot 2 \log_5 6\)

Now, \(\log_4 5 \cdot \log_5 6 = \log_4 6\) (by change of base identity)

Therefore, \(\log_4 y = 2 \log_4 6 = \log_4 6^2 = \log_4 36\)

Hence, \(y = 36\)