1. ## Monster problem

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

2. Originally Posted by Intrusion
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...

Very good! You can reduce that a little more
$2^{4x}- 2^{2(2x-2)}+ 2^{3x}= (2^x)^4- (2^{-4})(2^x)^4+ (2^x)^3= 4$

Now, let $y= 2^x$ and the equation becomes
$y^4- \frac{1}{16}y^4+ y^3= 4$
or $17y^4+ 16y^3= 4$

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