Question

Prove that : ${\log }_{5}1125=2{\log }_{5}6-\frac{1}{2}{\log }_{5}16+6{\log }_{49}7$

Answer 1

R.H.S.= $2{\log }_{5}6-\frac{1}{2}{\log }_{5}16+6{\log }_{49}7$

$={\log }_5{6}^2-{\log }_5{\left(16\right)}^{\frac{1}{2}}+3\times 2{\log }_{49}7={\log }_{5}36-{\log }_{5}4+3{\log }_{49}{7}^2$

$={\log }_5\left(\frac{36}{4}\right)+3{\log }_{49}49={\log }_{5}9+3\left(1\right)$

$={\log }_{5}9+3{\log }_{5}5={\log }_{5}9+{\log }_5{\left(5\right)}^3$

$={\log }_{5}9+{\log }_{5}125={\log }_5(9\times 125)={\log }_{5}1125$ = L.H.S.

$\therefore {\log }_{5}1125=2{\log }_{5}6-\frac{1}{2}{\log }_{5}16+6{\log }_{49}7$