Question

Prove that $\frac{1}{\log_{6}24} +\frac{1}{\log_{12}24}+\frac{1}{\log_{8}24}=2$

Answer 1

Please see the attachment

Answer 2

Statement to prove : $\frac{1}{\log_{6}24} +\frac{1}{\log_{12}24}+\frac{1}{\log_{8}24}=2$

Step by step proof :


$\frac{1}{\log_{6}24} +\frac{1}{\log_{12}24}+\frac{1}{\log_{8}24} \\ \\ = log_{24}(6) + log_{24}(12) + log_{24}(8) \\ \\ = log_{24}(6 \times 12 \times 8) \\ \\ = log_{24}(576) \\ \\ = log_{24}( {24}^{2} ) \\ \\ = 2$
Hence proved.

Identities used :
$log_{a}(b) = \frac{1}{ log_{b}(a) } \\ \\ log_{a}(m ^{n} ) = n \: \: log_{a}(m)$