Question

Show that ${\log }_{6}7=\frac{lo{g}_{2}7}{1+lo{g}_{2}3}$

Answer 1

Given,

${\log }_{6}7$

using the base property of log

$=\frac{\log 7}{\log 6}$

$=\frac{\log 7}{\log (2\times 3)}$

$=\frac{\log 7}{\log 2+\log 3}$

taking ${\log }_2$, we get,

$=\frac{{\log }_{2}7}{{\log }_{2}2+{\log }_{2}3}$

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

Henced proved.