If $a = {\log _{24}}12,\,b = {\log _{36}}24$ and $c = {\log _{48}}36,$ then $1+abc$ is equal to
If $a = {\log _{24}}12,\,b = {\log _{36}}24$ and $c = {\log _{48}}36,$ then $1+abc$ is equal to
- A
$2ab$
- B
$2ac$
- C
$2bc$
- D
$0$
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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
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- [KVPY 2021]
If ${\log _{0.04}}(x - 1) \ge {\log _{0.2}}(x - 1)$ then $x$ belongs to the interval
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The number ${\log _2}7$ is
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