16^x - 4^(2x-2) + 2^(3x) = 4
I've gotten to:
2^4x - 2^2(2x-2) + 2^3x = 2^2
I've gotten beyond lost after here...
I've also tried dividing by 2^2 to get....
2^(4x-2) - 2^(4x-6) + 2^(3x-2) = 1 but I'm not sure what I can do with this
Very good! You can reduce that a little more
$\displaystyle 2^{4x}- 2^{2(2x-2)}+ 2^{3x}= (2^x)^4- (2^{-4})(2^x)^4+ (2^x)^3= 4$
Now, let $\displaystyle y= 2^x$ and the equation becomes
$\displaystyle y^4- \frac{1}{16}y^4+ y^3= 4$
or $\displaystyle 17y^4+ 16y^3= 4$
but, frankly, I don't see any simple solution to that.
I've also tried dividing by 2^2 to get....
2^(4x-2) - 2^(4x-6) + 2^(3x-2) = 1 but I'm not sure what I can do with this