Powers of 2 and 3

Since we know $2^{x} = 9$ , we can rewrite that as $\log_{2}9 = x$ . Also, since $27^{y} = 2$ we know $3^{3y} = 2$ which can be rewrite as $\log_{3}2 = 3y$ . Now, we see that we should convert the first equation to base three, which gives us $\frac{\log_{3}9}{\log_{3}2}$ By multiplying $3y$ and $x$ , we actually multiply $\log_{3}2$ and $\frac{\log_{3}9}{\log_{3}2}$ , leaving us with $\log_{3}9 = 2$

taking logrithm on both sides for both eqns.. we get x= (2log3)/log2 and y=log2/(3log3) multiplying x and y we get x*y= 2/3 therefore 3xy= 3x2/3 = 2 answer...

2^x=9

2^x=3^2

2=3^(2/x)

27=2

3^3y=2

3^3y=3^(2/x)

3y=2/x

3yx=2