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

I was doing the problem

Find all real solutions for $$x$$ in:

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

There was a hint, to prove that $$2^{x} - 1$$ has the same sign as $$x$$, although with basic math, if $$2^0 - 1$$ = 0, and x in this case = 0, The two expressions will have the same sign, so I am just puzzled on where to go next with this problem, any help would be welcome!

Messing around with basic values, I got $$\pm 1, 0$$ as the answers, although I have not checked for extra solutions, and do not know how to prove these, as it was just basic guessing and checking with the three most basic numbers for solving equations.

Now, for $$x\neq1$$, $$x\neq0$$ and $$x\neq-1$$ rewrite our equation in the following form: $$\frac{2^x-1}{x}+\frac{2^{x^2-1}-1}{x^2-1}=0.$$ Can you end it now?
Dividing both sides by $$2$$, we get $$x^2(2^x-1)+x(2^{x^2-1}-1)=2^x-1$$. Subtracting $$2^x-1$$ from both sides gives $$(x^2-1)(2^x-1)+x(2^{x^2-1}-1)=0$$. We can also note that for any $$x>1$$, all terms on the LHS of the equation $$(x^2-1)(2^x-1)+x(2^{x^2-1}-1)=0$$ are positive. For all $$x<-1$$, $$x^2-1$$ is positive, $$2^x-1$$ is negative, $$x$$ is negative, and $$2^{x^2-1}-1$$ is positive, so the whole expression is negative. Therefore, our solutions must be in the range $$-1 \leq x \leq 1$$.