Find al real solutions of $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2$

Question

Not sure how to do this. I had a hint to prove that $(2^x- 1)$ and $x$ have the same sign, but how would that help me? Please help!

Answer 1

Hint:

$$(x^2-1)(2^{x+1}-2)=-(2^{x^2}-2)x$$

Answer 2

Initial equation can be reduced to $(2^{x}-1)(x^{2}-1) + x(2^{x^{2}-1}-1)=0$. Let $y = (x^{2}-1)$, then equation transforms to $(2^{x}-1)y + x(2^{y} - 1)=0$. If you can prove that $(2^{x}-1)$ and $x$ have the same sign then you can see that $(2^{x}-1)y$ and $x(2^{y} - 1)$ also have the same sign. Which means that the only solutions are $x = 0$ and $y = 0$ (that is $x^{2}-1=0$)