2^(x^2 - 35)= 4^x

How do I solve this? The x^2 is messing me up. So far I have:
ln(2^(x^2-35))=ln(4^x)
(x^2-35)ln 2= xln4
x^2ln 2 - 35ln2= xln4
x^2ln 2- xln4= 35ln2
x(xln2 - ln4)= 35ln2

I would have taken the log base 2 of each side, but ok, with what you have, ln4 is 2ln2. Then divide through by ln2.

here is an easy way to do that.
Notice the bases of the powers are 2 and 4, so let's change 4 to base 2

2^(x^2 - 35)= 4^x
2^(x^2 - 35)= 2^2x
then x^2 - 35 = 2x
x^2 - 2x - 35 = 0
(x-7)(x+5)=0
x=7 or x=-5