Question

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

Answer 1

$LHS=\frac{1}{lo{g}_{6}24}+\frac{1}{lo{g}_{12}24}+\frac{1}{lo{g}_{8}24}$

Changing base to $2,$ by change of base theorem

$=\frac{1}{\frac{lo{g}_{2}24}{lo{g}_{2}6}}+\frac{1}{\frac{lo{g}_{2}24}{lo{g}_{2}12}}+\frac{1}{\frac{lo{g}_{2}24}{lo{g}_{2}8}}$

$=\frac{lo{g}_{2}6+lo{g}_{2}12+lo{g}_{2}8}{lo{g}_{2}24}$

$=\frac{lo{g}_2(6\times 12\times 8)}{lo{g}_{2}24}$

$=\frac{lo{g}_{2}576}{lo{g}_{2}24}=\frac{lo{g}_2{24}^2}{lo{g}_{2}24}$

$=2\frac{lo{g}_{2}24}{lo{g}_{2}24}$ property $(log{b}^m=mlogb)$

$=2$

$=R.H.S.$