Question

Show that :
$\frac{1}{lo{g}_{2}42}+\frac{1}{lo{g}_{3}42}+\frac{1}{lo{g}_{7}42}=1$

Answer 1

To prove:- $\frac{1}{{\log }_{2}42}+\frac{1}{{\log }_{3}42}+\frac{1}{{\log }_{7}42}=1$

Proof:-

Taking L.H.S.-

$\frac{1}{{\log }_{2}42}+\frac{1}{{\log }_{3}42}+\frac{1}{{\log }_{7}42}$

$={\log }_{42}2+{\log }_{42}3+{\log }_{42}7(\because {\log }_{a}b=\frac{1}{{\log }_{b}a})$

$={\log }_{42}(2\times 3)+{\log }_{42}7$

$={\log }_{42}6\times 7$

$={\log }_{42}42$

$=1$

$=$R.H.S.

Hence proved that $\frac{1}{{\log }_{2}42}+\frac{1}{{\log }_{3}42}+\frac{1}{{\log }_{7}42}=1$