4^2014-4^2013-4^2012+4^2011=45(2^x) solving for x.

Despite what everyone thinks you can divide exponents with like bases, it's not just dividing. this is my work so far: i factored out 4^2012 first
4^2012(4^2-4^1-4^0)+4^2011=45(2^x)
4^2012(16-4-1)+4^2011=45(2^x)
4^2012(11)+4^2011=45(2^x)
then factored out 4^2011
4^2011(11*4^1+4^0) <----this equals the previous line... 4^2012(11)+4^2011=45(2x)
4^2011(45)=45(2^x)
4^2011=2^x
this is obviously a VERY large number, and there's no way of rounding since it's an integer with no decimals, but there's got to be some type of answer!! does anyone have any ideas?

Assuming your math is correct up to that point, you're practically done.

4^2011 = 2^x

Since 4 = 2^2, then [2^2]^2011 = 2^x

2^(2*2011) = 2^4022 = 2^x

x = 4022

Edit: By the way, you could have just factored out 4^2011 in the first step.
4^2011*(4^3 - 4^2 - 4 + 1) = 4^2011*(64 - 16 - 4 + 1) = 4^2011*45

45 * 2^x = 4^(2011) * (1 - 4 - 16 + 64)
45 * 2^x = 2^(4022) * (45)
x = 4022

all i see is 420 420 420 420 420 420 420 lollll