Log - log-ab-a-1-log-b-ab-01log abNone of these

The correct option is A 0log { log_ab a+1/log_b ab}= = log { log_ab a+log_ab b } = log (log_ab ab) =log 1=0 ...

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Fundamental laws of logarithms

log { log_ab ...

Question

$l o g {l o g_{a b} a + \frac{1}{l o g_{b} a b}} =$

0.0

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log ab

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None of these

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l o g {l o g_{a b} a + \frac{1}{l o g_{b} a b}} =

2^{(\sqrt{l o g_{a} \sqrt[4]{a b} + l o g_{b} \sqrt[4]{a b}} - \sqrt{l o g_{a} \sqrt[4]{b / a} + l o g_{b} \sqrt[4]{a / b}}) . \sqrt{l o g_{a} b}}

= {\begin{matrix} 2, b \geq a \geq 1 2^{l o g_{a} b}, 1 < b < a \end{matrix}

\int_{0}^{1} \log (\frac{1}{x} - 1) d x

a > b

a^{b} b^{b} > a^{b} b^{a}

log \frac{b}{a} < log \frac{1 + b}{1 + a}

\int_{0}^{1} log [\sqrt{1 - x} + \sqrt{1 + x}] d x = \frac{1}{2} log 2 + \frac{π}{k} - \frac{1}{2}

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