Question

Please help me with this

Answer 1

Answer

$\bullet \;\;\tt \log_{a^2}a.log_{b^2}b.log_{c^2}c=\dfrac{1}{8}$

Given

$\bullet \;\; \tt \log _{a^2}a.\log_{b^2}b.\log_{c^2}c$

To Find

$\bullet \tt \;\; \dfrac{1}{8}$

Formula Used

$\tt \bullet \;\; \log_xy=\dfrac{\log y}{\log x}\\\\\bullet \;\; \tt \log x^y=y.\log x$

Solution

$\tt LHS\\\\\implies \tt \log _{a^2}a.log_{b^2}b.log_{c^2}c\\\\\implies \tt \dfrac{\log {a}}{\log a^2}.\dfrac{\log b}{\log b^2}.\dfrac{\log c}{\log c^2}[By\ Formula\ 1]\\\\\implies \tt \dfrac{\log a}{2 \log a}.\dfrac{\log b}{2 \log b}.\dfrac{\log c}{2 \log c}[By\ Formula\ 2]\\\\\implies \tt \dfrac{\cancel{\log a}}{2\ \cancel{\log a}}.\dfrac{\cancel{\log b}}{2\ \cancel{\log b}}.\dfrac{\cancel{\log c}}{2\ \cancel{\log c}}\\\\\implies \tt \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}$

$\implies \bf \dfrac{1}{8}\\\\\implies \tt RHS$

Hence Proved