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

If the domain of \[ f(x)=\log_{(10x^2-17x+7)}\,(18x^2-11x+1) \] is $(-\infty,a)\cup(b,c)\cup(d,\infty)-\{e\}$, then find $90(a+b+c+d+e)$.

Answer 1

To determine $y$ from the equation:

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

First, simplify $ \log_6 \sqrt{5} $:

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

Substitute this back into the original equation:

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

Rearrange the equation to isolate $ \log_4 y $:

$ \log_4 y = \frac{2 \log_4 5}{\log_6 5} $

Convert the logarithms to base 10 for easier manipulation:

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

Substitute these base-10 expressions into the equation for $ \log_4 y $:

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

Cancel the common term $ \log_{10} 5 $:

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

Apply the change of base formula to express this in terms of $ \log_{10} y $:

$ \log_{10} y = \log_{10} 4 \times \log_4 y = \log_{10} 4 \times \frac{2 \log_{10} 6}{\log_{10} 4} = 2 \log_{10} 6 $

Using the property $ a \log b = \log b^a $, we get:

$ \log_{10} y = \log_{10} 6^2 = \log_{10} 36 $

Therefore, $ y = 36 $.

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