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Basic of Logarithms
hard
Let $a, b, x$ be positive real numbers with $a \neq 1$, $x \neq 1$, ab $\neq 1$. Suppose $\log _{ a } b =10$, and $\frac{\log _{ a } x \log _{ x }\left(\frac{ b }{ a }\right)}{\log _{ x } b \log _{ ab } x }=\frac{ p }{ q }$, where $p$ and $q$ are positive integers which are coprime. Then $p+q$ is
A
$9$
B
$99$
C
$109$
D
$199$
(KVPY-2021)
Solution
(c)
$\frac{p}{q}=\frac{\log _a x \cdot \log _x\left(\frac{b}{a}\right)}{\log _x b \cdot \log _{a b} x}$
$\frac{ p }{ q }=\frac{\frac{(\log b -\log a)}{\log a}}{\frac{\log b }{(\log a+\log b)}}$$\frac{ p }{ q }=\frac{(\log b )^2-(\log a)^2}{(\log a)(\log b )}$
Given $\log _{ a } b =10 \Rightarrow \log b =10 \log a$
$\frac{ p }{ q }=\frac{(100-1)(\log a)^2}{10(\log a)^2}=\frac{99}{10}$
$p+q=109$
Standard 11
Mathematics
