Question: Find all real solutions for $x$ in $$ 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 . $$
Hint: First prove that $2^x - 1$ and $x$ have the same sign.
My Approach: I proved the same sign thing, but I do not understand where to go from there. I tried breaking the equation down with $2(2^x-1)x^2+(2^2x-2)x=2(2^x-1).$ Then I substituted $2^x-1$ with $y,$ but I don't know where to go from there.
Please help.
The equation is equivalent to: $$(2^x- 1)(x^2-1) + (2^{x^2-1}-1)x=0$$ It is satisfied by $x\in \{-1,0,1\}$.
Moreover if $x\not\in \{-1,0,1\}$ then it can be written as $f(x)+f(x^2-1)=0$ with $f(x):=(2^x-1)/x$. Since $f(x)>0$ for any $x\not=0$ then $f(x)+f(x^2-1)>0$ for any $x\not\in \{-1,0,1\}$.
Hence the complete set of solutions is $\{-1,0,1\}$.