Question 481432
First, note that both bases are perfect squares: {{{16=4^2}}} and {{{9=3^2}}}.  Let's rewrite your equation:
{{{16^log3 = 9^log4}}}
{{{(4^2)^log3=(3^2)^log4}}}
Recall that when you raise a power to a power, you multiply the exponents:
{{{4^(2*log3)=3^(2*log4)}}}
We are allowed to take the logarithm of both sides of the equation:
{{{log4^(2*log3)=log3^(2*log4)}}}
My favorite property of logarithms is that you can "kick the exponent to the front" :)
{{{2*log3*log4=2*log4*log3}}}
Rearrange to make it obvious (commutative property):
{{{2*log3*log4=2*log3*log4}}}