Solve
4^2x=2^5-x
and
9^x+1=27^2x-3
For the first one I keep getting like .4, and forthe life of me I can not figure out what im not doing right.
I assume you mean $\displaystyle 4^{2x} = 2^{5-x}$
Note that $\displaystyle 4^{2x} = 2^{4x}$
Now if two bases are equal then their exponents must also be equal so you get the linear equation $\displaystyle 4x = 5-x$ which is easy to solve
For question 2 use the fact that $\displaystyle 9 = 3^2 \text{ and } 27 = 3^3$