Question

If ${\log }_{3}m=x$ and ${\log }_{3}n=y$, then $3^{1-2y+3x}$ can be expressed in terms of $m$ and $n$ as $\frac{3m^3}{n^2}$.

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

Answer 1

Given, ${\log }_{3}m=x$ and ${\log }_{3}n=y$
$\therefore m=3^x$ and $n=3^y$ ...(1)
Thus, $3^{1-2y+3x}$$=\frac{3\times 3^{3x}}{3^{2y}}$

$\therefore \frac{3\times 3^{3x}}{3^{2y}}=\frac{3m^3}{n^2}$ (From 1)