Question

If \(log_45=(log_4y)(log_6\sqrt5)\),then \(y\) equals
[This Question was asked as TITA]

• A. 34
• B. 36
• C. 38
• D. 37

Answer 1

The Correct Option is B

To find \(y\), given the equation:

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

First, express \( \log_6 \sqrt{5} \):

\( \log_6 \sqrt{5} = \log_6 5^{1/2} = \frac{1}{2} \log_6 5 \)

Substitute back into the equation:

\( \log_4 5 = (\log_4 y) \left(\frac{1}{2} \log_6 5\right) \)

Rearrange the formula:

\( \log_4 5 = \frac{1}{2} (\log_4 y) (\log_6 5) \)

Solve for \( \log_4 y \):

\( \log_4 y = \frac{2 \log_4 5}{\log_6 5} \)

Rewrite logs to base 10:

\( \log_4 5 = \frac{\log_{10} 5}{\log_{10} 4} \) and \( \log_6 5 = \frac{\log_{10} 5}{\log_{10} 6} \)

Substitute these into the equation:

\( \log_4 y = \frac{2 \left(\frac{\log_{10} 5}{\log_{10} 4}\right)}{\frac{\log_{10} 5}{\log_{10} 6}} \)

Cancel out \( \log_{10} 5 \):

\( \log_4 y = \frac{2 \log_{10} 6}{\log_{10} 4} \)

Convert to base 10 log:

\( \log_{10} y = \frac{\log_{10} 6}{\log_{10} 4} \times 2 \) (using change of base formula)

Recall \(\log_{10} 6 \approx 0.778\) and \(\log_{10} 4 \approx 0.602\):

\( \log_{10} y \approx \frac{0.778}{0.602} \times 2 \approx 1.079 \times 2 \approx 2.158 \)

Thus, \( y \approx 10^{2.158} \).

Calculating \( y \):

\( 10^{2.158} \approx 36 \).

Therefore, \( y \) is 36.

Answer 2

Given:
\(\log_45=(\log_4y)(log_6\sqrt5)\)
Now we use the change of base formula which is \(\log_ab=\frac{\log b}{\log a}\)
\(\log_6\sqrt5=\frac{\log\sqrt5}{\log 6}\)

Since \(\log\sqrt5-\frac{1}{2}\log5,\) we have:

\(\log_6\sqrt5=\frac{1}{2}\cdot\frac{\log5}{\log6}\)
Then, our equation becomes:
\(\log_45=(\log_4y)(\frac{1}{2}\cdot\frac{\log 5}{\log 6})\)

We again use the change of base formula, so \(\log_45=\frac{\log5}{\log4}\), let's substitute this into the equation:
\(\frac{\log5}{\log4}=(\log_4y)(\frac{1}{2}\cdot\frac{\log5}{\log6})\)

Now, lets simplify this equation:
\(\frac{\log5}{\log4}=\frac{\log5}{2\log6}\cdot\log_4y\)

Cancel out log5 from both sides, the we get:
\(\frac{1}{\log4}=\frac{1}{2\log6}\cdot\log_4y\)

\(\log_4y=\frac{\log4}{2\log6}\)

Now we use the power property of logarithms \(\log_b(x^y)=y\cdot\log_b(x)\)
\(\log_4y=\frac{\log4}{2\log6}\)

\(\log_4y=\frac{\log4}{\log6^2}=\frac{\log4}{\log36}\)

\(\log_4y=\frac{\log4}{\log36}\), as we see in the change of base formula: \(\log_ab=\frac{\log b}{\log a}\)
y=36

So, the correct option is (B): 36.