Problem:
Find all real solutions for $x$ in $$ 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 . $$
I know that you can move all the terms to one side, then divide by $2$ and factor, but what next?
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JenkinsMa
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3$$(2^x-1)(x^2-1)+x(2^{x^2-1}-1)=0\implies x^2-1=0$$ is a solution – lab bhattacharjee Apr 15 '17 at 15:53
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If I may ask, how did you get from the original equation to $x^2-1=0?$ – JenkinsMa Apr 15 '17 at 15:54
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2$x=0$ is one solution. – JJR Apr 15 '17 at 16:04
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1there are three real solutions – Dr. Sonnhard Graubner Apr 15 '17 at 16:23
1 Answers
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We seek all real solutions for $2(2^x-1)x^2+\left(2^{x^2}-2\right)x=2^{x+1}-2$.
Let's rewrite our equation into a very interesting form.
\begin{align} 2(2^x-1)x^2+\left(2^{x^2}-2\right)x&=2^{x+1}-2\\ 2(2^x-1)x^2+\left(2^{x^2}-2\right)x-2^{x+1}+2&=0&&\text{Set equation = 0}\\ (2^x-1)x^2+\left(2^{x^2-1}-1\right)x-2^x+1&=0&&\text{Divide all terms by 2}\\ (2^x-1)x^2+\left(2^{x^2-1}-1\right)x-(2^x-1)&=0&&\text{Factor out $-1$}\\ (2^x-1)(x^2-1)+\left(2^{x^2-1}-1\right)x&=0&&\text{Factor out $2^x-1$}\\ (2^x-1)y+x(2^y-1)&=0\tag{1}\label{eq}&&\text{Replace $x^2-1$ with $y$} \end{align}
Notice the symmetry of the left-hand side of Equation \ref{eq}. Also notice that if $x=0$, then $2^x-1=0$ as well.
The solutions of Equation \ref{eq} must be values such that $x=0$ or $y=x^2-1=0$. Therefore the solutions are $x=0, 1, -1$.
Tim Thayer
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