Question

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

Answer 1

Taking LHS

$\frac{1}{{\log }_{x}y}={\log }_{y}x$

Therefore the equation becomes

${\log }_{24}6+{\log }_{24}12+{\log }_{24}8$

We know that ${\log }_{z}a+{\log }_{z}b+{\log }_{z}c=lo{g}_z\left(a\times b\times c\right)$

Therefore the equation becomes

${\log }_{24}\left(6\times 12\times 8\right)$

${\log }_{24}576$

${\log }_{24}(24{)}^2$...............(1)

We know that ${\log }_b{a}^2=2\times {\log }_{b}a$

Therefore the equation (1) becomes

$2\times lo{g}_{24}24$..................... (${\log }_{a}a=1$)

$2=RHS$

Hence proved

$RHS=LHS$