The value of is equal to

JEE MainMathematicsEasyLogarithm and its applications

The value of
$\frac{1}{\log_{\sqrt{bc}}(abc)} + \frac{1}{\log_{\sqrt{ca}}abc} + \frac{1}{\log_{\sqrt{ab}}(abc)}$
is equal to

Show answer and solution

\frac{1}{\log_{\sqrt{bC}}(abc)} + \frac{1}{\log_{\sqrt{ca}}{abc}\ } + \frac{1}{\log_{\sqrt{ab}}{(abc)}}

= \log_{abc}\sqrt{bc} + \log_{abc}\sqrt{ca} + \log_{abc}\sqrt{ab}

= \frac{1}{2}\left( \log_{abc}bc + \log_{abc}ca + \log_{abc}ab \right)

= \frac{1}{2}\left( \log_{abc}(bc \times ca \times ab) \right)

= \frac{1}{2}\left( \log_{abc}(abc)^{2} \right)

= \frac{1}{2} \times 2\log_{abc}abc

= 1

Source: Grass-G / M-05 Mathematics Paper 10 October.docx

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