Question

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

Answer 1

\(log_45=(log_4y)(log_6\sqrt5)\)

\(⇒ \frac{log_45}{log_6\sqrt5} = log_4y\)

\(⇒ log_45×log_6\sqrt5 = log_4y\)

\(⇒ 2(log_45)(log_56) = (log_4y)\)

\(⇒ 2log_46 = log_4y\)

\(⇒ log_46^2 = log_4y\)

\(⇒ log_436 = log_4y\)

\(⇒ y = 36\)

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 answer is 36.