1.$\displaystyle \log_{2}(25^(x+3) - 1) = 2 + \log_{2}(5 ^ (x+3) + 1)$

2.$\displaystyle \log_{4}(2 \cdot 3^(x) - 6) -\log_{2}(9^(x) - 6)$

3.$\displaystyle \log_{4}(2 \cdot 4^(x-2) - 1) + 4 = 2x$

I attempted the first to come as far as this-:

Taking x+3 = n,

$\displaystyle 25^(n) - 1/5^(n) + 1 = 4$

I've forgotten exponential mathematics it seems..

Pardon me if I seem like a total noob, but I need help :P