1.Solve, 3^2x = 5^(1-x)
2. Solve 4^2x - 8(2^2x) - 9 =0
Hm... let's do it.
$\displaystyle 3^{2x} = 5^{1-x}$
Take logs on both sides.
$\displaystyle log3^{2x} = log5^{1-x}$
$\displaystyle 2x log3 = (1-x)log5$
$\displaystyle 2x log3 = log5 - xlog5$
Bring xlog5 to the left;
$\displaystyle 2x log3 + x log 5 = log5$
Factor x;
$\displaystyle x (2log3 + log 5) = log5$
Can you continue now?