Question

Given : ${\log }_{3}m=x$ and ${\log }_{3}n=y$

If $2{\log }_{3}A=5x-3y$, then $A=\sqrt{\frac{m^5}{n^3}}$.

If true then write 1 and if false then write 0.

Answer 1

Given, ${\log }_{3}m=x,{\log }_{3}n=y$

Also, $2{\log }_{3}A=5x-3y$

$⟹2{\log }_{3}A=5{\log }_{3}m-3{\log }_{3}n$

$⟹2{\log }_{3}A={\log }_3{m}^5-{\log }_3{n}^3$

$⟹{\log }_{3}A={\log }_3\sqrt{\frac{m^5}{n^3}}$

$A=\sqrt{\frac{m^5}{n^3}}$.