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Find all real solutions for $x$ in $2(2^x−1)x^2+(2^{x^2}−2)x=2^{x+1}−2$.

I started off by dividing $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2$ by $2$, and I got $(2^x-1)x^2 + (2^{x^2-1}-1)x = 2^x -1$. I tried dividing by $2^x-1$ on both sides which would give me a simple quadratic to solve for $x$, but I don't know how to simplify $\frac{(2^{x^2}-1)}{(2^x-1)}$. Am I missing a simple trick here, or am I on a completely wrong path to solving this?

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  • $\begingroup$ $x = 0$ is clearly a solution, and the left hand side is increasing much more rapidly than the right. You can probably use this to get a lot of information. $\endgroup$
    – user296602
    Mar 5 '18 at 3:35
  • $\begingroup$ You cannot simplify $\dfrac{2^{x^2-1}-1}{2^x-1}$. $\endgroup$
    – user535339
    Mar 5 '18 at 3:43
  • $\begingroup$ -1 seems to be a solution too $\endgroup$
    – ir7
    Mar 5 '18 at 3:52
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    $\begingroup$ Hint: write it as $\;2(2^x−1)(x^2-1)+(2^{x^2}−2)x=0\,$, then the solutions $\,0, \pm 1\,$ become more obvious. Now break it down by intervals, for example if $\,x \in (0,1)\,$ then the LHS is strictly negative etc. $\endgroup$
    – dxiv
    Mar 5 '18 at 3:52
  • $\begingroup$ And so is $1$.. $\endgroup$
    – user535339
    Mar 5 '18 at 3:52
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As per dxiv's hint: by elementary methods, the equation is equivalent to

$$f(x)\stackrel{\text{def}}{=}(2^x−1)(x^2-1)+(2^{x^2-1}−1)x=0$$ Note that, for any $y$, $2^y-1$ and $y$ have the same sign (positive, negative, or zero); write $\mathrm{sgn}\,y$ for the sign of $y$. Then

$$\mathrm{sgn}[(2^x-1)(x^2-1)]=\mathrm{sgn}(2^x-1)\mathrm{sgn}(x^2-1)=\mathrm{sgn}(x)\mathrm{sgn}(x^2-1)$$ and likewise $$\mathrm{sgn}[(2^{x^2-1}-1)x]=\mathrm{sgn}(x^2-1)\mathrm{sgn}(x)$$

Now, if $\mathrm{sgn}(y)=\mathrm{sgn}(z)$, then $\mathrm{sgn}(y+z)=\mathrm{sgn}(y)=\mathrm{sgn}(z)$. Consequently $$\mathrm{sgn}[f(x)]=\mathrm{sgn}(x)\mathrm{sgn}(x^2-1)=\mathrm{sgn}(x)\mathrm{sgn}(x+1)\mathrm{sgn}(x-1)\text{.}$$ By the zero product property and the fact that $\mathrm{sgn}(y)=0$ iff $y=0$, we find that $f(x)=0$ iff $x\in\{-1,0,1\}$.

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Using Mathematica

eq = 2 (2^x - 1) x^2 + (2^(x^2) - 2) x == 2^(x + 1) - 2;

sol = Solve[eq, x, Reals]

(* {{x -> -1}, {x -> 0}, {x -> 1}} *)

Verifying the solutions

And @@ (eq /. sol)

(* True *)

Showing the solutions on a plot

Plot[{2 (2^x - 1) x^2 + (2^(x^2) - 2) x, 2^(x + 1) - 2},
 {x, -2, 2}, PlotLegends -> Placed["Expressions", {.22, .75}],
 Epilog -> {Red, AbsolutePointSize[5],
   Point[{x, 2^(x + 1) - 2} /. sol]}]

enter image description here

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  • $\begingroup$ This provides a good hint, indeed, but a graph is still not a proof. $\endgroup$
    – dxiv
    Mar 5 '18 at 4:44
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$\begin{align*} 2\left (2^{x}-1 \right )x^{2}+\left(2^{x^{2}}-2\right)x&=2^{x}.2-2\\ 2\left(2^{x}-1\right)x^{2}-\left(2^{x^{2}}-2\right)x&=2(2^{x}-1) \end{align*}$

Let $2^{x}-1=a$ and $2^{x^{2}}-2=b$ and equation becomes: $2ax^{2} + bx=2a$ or $2ax^{2}+bx-2a=0$. It's a quadratic equation. The quadratic equation has real solution if and only if $b^{2}-4ac>0$.

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    $\begingroup$ Who on earth upvoted this answer within 15 seconds of being posted, and before the first edit, which made the MathJax even readable at all? $\endgroup$
    – dxiv
    Mar 5 '18 at 3:55
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    $\begingroup$ Kind of confused as how you treat variables as coefficients of quadratic $\endgroup$
    – King Tut
    Mar 5 '18 at 3:55
  • $\begingroup$ @KingTut, there's nothing wrong with a quadratic-formula-based argument in which the coefficients are themselves functions of the variable. The problem here is that the discriminant for $2ax^2+bx-2a$ is $b^2+8a^2$, which is clearly always non-negative, so there's nothing gained from the approach. $\endgroup$ Mar 5 '18 at 16:48
  • $\begingroup$ @barry i understand what you are implying, but had it been someother variable say Z then this approach would have been useful. coefficients are themselves function of the same variable x. So to me this approach seemed ridiculous. I may be wrong.. yes and barry i understand your point about discriminant.. $\endgroup$
    – King Tut
    Mar 5 '18 at 17:08
  • $\begingroup$ @KingTut, I don't recall when or why, but I do recall once making an argument along these lines, in which the existence of real solutions of an equation of the form $a(x)x^2+b(x)x+c(x)=0$ depended on $b^2-4ac$ being non-negative. So the approach doesn't strike me as inherently unworkable. It just doesn't seem to work in this case. $\endgroup$ Mar 5 '18 at 17:17

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