Let a, b, x be positive real numbers with a n | vedclass

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(C) Given $\log_{a} b = 10$,which implies $\log b = 10 \log a$.The expression is $\frac{\log_{a} x \cdot \log_{x}(\frac{b}{a})}{\log_{x} b \cdot \log_

Vedclass · Accepted Answer

$109$

Vedclass · Answer

(C) Given $\log_{a} b = 10$,which implies $\log b = 10 \log a$.The expression is $\frac{\log_{a} x \cdot \log_{x}(\frac{b}{a})}{\log_{x} b \cdot \log_{ab} x}$.Using the change of base formula $\log_{m} n = \frac{\log n}{\log m}$,we have:$\log_{a} x = \frac{\log x}{\log a}$,$\log_{x}(\frac{b}{a}) = \frac{\log b - \log a}{\log x}$,$\log_{x} b = \frac{\log b}{\log x}$,and $\log_{ab} x = \frac{\log x}{\log a + \log b}$.Substituting these into the expression:$\frac{p}{q} = \frac{(\frac{\log x}{\log a}) \cdot (\frac{\log b - \log a}{\log x})}{(\frac{\log b}{\log x}) \cdot (\frac{\log x}{\log a + \log b})} = \frac{\frac{\log b - \log a}{\log a}}{\frac{\log b}{\log a + \log b}} = \frac{(\log b - \log a)(\log a + \log b)}{\log a \cdot \log b} = \frac{(\log b)^2 - (\log a)^2}{\log a \cdot \log b}$.Since $\log b = 10 \log a$,we substitute this into the expression:$\frac{p}{q} = \frac{(10 \log a)^2 - (\log a)^2}{\log a \cdot (10 \log a)} = \frac{100(\log a)^2 - (\log a)^2}{10(\log a)^2} = \frac{99(\log a)^2}{10(\log a)^2} = \frac{99}{10}$.Since $p=99$ and $q=10$ are coprime,$p+q = 99+10 = 109$.

Let $a, b, x$ be positive real numbers with $a \neq 1, x \neq 1, ab \neq 1$. Suppose $\log_{a} b = 10$,and $\frac{\log_{a} x \cdot \log_{x}(\frac{b}{a})}{\log_{x} b \cdot \log_{ab} x} = \frac{p}{q}$,where $p$ and $q$ are positive integers which are coprime. Then $p+q$ is

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