How to prove solutions of $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2$?

Question

The original question wanted the real solutions to $$ 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 . $$ I graphed this equation and worked out the trivial roots $x=-1$ and $x=1$, and I could not find any more. Checking again, these are the only two solutions. So the problem is, I don't know how to write a formal derivation of those roots.

Thank you in advance for your help.

Answer 1

Since for the right-hand term of the original equation we have $2^{x+1}-2=2\left(2^x-1\right)$, the equation can be rewritten as: $$2\left(2^x-1\right)\left(x^2-1\right)+2\left(2^{x^2-1}-1\right)x=0,$$ which after simplifying the factor $2$ reduces to the even simpler form: $$\left(2^x-1\right)\left(x^2-1\right)+\left(2^{x^2-1}-1\right)x=0.$$ It is easy to ascertain that the roots of the equations $x=0$ respectively $x^2-1=0$ (in other words, the elements of the set $\{-1, 0, 1\}$) are also roots of our equation. We proceed to show they are the only roots, by assuming the existence of an additional real root $a \in \mathbb{R} \setminus \{-1, 0, 1\}$ of our equation, other than these special values. In this case, via dividing by $a\left(a^2-1\right)\neq0$ the equation can be brought to the form: $$\frac{2^a-1}{a}+\frac{2^{a^2-1}-1}{a^2-1}=0.$$ We remark that for any $y \in \mathbb{R}^{\times}=\mathbb{R} \setminus \{0\}$ and and any $t \in (1, \infty)$ the inequality $\frac{t^y-1}{y}>0$ holds, so by the particular application of this observation for $t=2$ and $y=a$ respectively $y=a^2-1$ we gather that the left-hand side of the last equation above is strictly positive, which constitutes a contradiction. This concludes the argument.

P.S. There is a very natural way to immediately generalise this exercise and for this we require the following ingredients (objects):

• a commutative totally ordered field $(K, +, \cdot, R)$, in other words a commutative field $(K, +, \cdot)$ where the additional entity $R$ is a total order on the support set $K$ that is also compatible with the operations. Since the order is total, any element $x \in K$ satisfies only one of the relations $x<_R 0_K$, $x=0_K$, $x>_R 0_K$ and we go on to define the sign function: \begin{align*} \mathrm{sgn} \colon K &\to \left\{-1_K, 0_K, 1_K\right\} \\ \mathrm{sgn}(x)&=\begin{cases} -1_K, &x<_R 0_K \\ 0_K, &x=0_K \\ 1_K, &x>_R 0_K. \end{cases} \end{align*}
• for any arbitrary set $M$ we introduce the abbreviation $\mathrm{T}(M)\colon=\mathrm{End}_{\mathbf{Ens}}(M)$ for the collection of all self-maps of $M$ (which carries a natural monoid structure with respect to composition of maps, the so-called transformation monoid of set $M$). Given arbitrary self-map $f \in \mathrm{T}(M)$ we say subset $X \subseteq M$ is stable under $f$ if $f[X] \subseteq X$ and we write: $$\mathscr{S}\mathscr{tab}(f)\colon=\{X \subseteq M \mid f[X] \subseteq X\}$$ for the collection of all subsets stable under $f$. Returning to the commutative totally ordered field introduced above, let us write: $$\mathrm{S}_R(K)\colon=\{f \in \mathrm{T}(K) \mid (\leftarrow, 0_K)_R, \{0_K\}, (0_K, \rightarrow)_R \in \mathscr{Stab}(f)\}$$ for the collection of all sign-preserving self-maps of $K$ (where of course the sign is determined with respect to the fixed order relation $R$). More explicitly, a sign-preserving map will map $0_K$ to itself respectively strictly positives (negatives) to themselves. Note the equivalence: $$f \in \mathrm{S}_R(K) \Leftrightarrow f \in \mathrm{T}(K) \wedge \mathrm{sgn} \circ f=\mathrm{sgn}.$$
• arbitrary sets $A$ and $I\neq \varnothing$, where the latter is additionally required to be finite. Furthemore consider two families of maps, $f \in \mathrm{Hom}_{\mathbf{Ens}}(A, K)^I$ (this is simply a family of maps from $A$ to $K$ indexed by $I$) and $g \in \mathrm{S}_R(K)^I$ (this being a family of sign-preserving maps of $K$, again indexed by the same $I$).

If we define the map: $$h=\sum_{i \in I}\left(\left(g_i \circ f_i\right)\prod_{j \in I \setminus \{i\}}f_j\right),$$ then the roots of $h$ are described by the relation: $$h^{-1}[\{0_K\}]=\displaystyle\bigcup_{i \in I}f_i^{-1}[\{0_K\}].$$ In the particular case of your exercise we have $K=\mathbb{R}$ with the standard ordered field structure, $A=\mathbb{R}$, $I=\{1, 2\}$, $g_1=g_2$ given by $g_1(x)=2^x-1$ (in general the map $x \mapsto a^x-1$ will be sign-preserving as long as $a>1$) and finally $f_1=\mathbf{1}_{\mathbb{R}}$ respectively $f_2(x)=x^2-1$ for all $x \in \mathbb{R}$.

Answer 2

Partial solution: Here is an approach which didn't really pan out completly:

Rewrite as : $$2^{x}(x^2 - 1) + x2^{x^2 - 1} = x^2 + x - 1 $$

Let $y = x^2 - 1$, then we get: $$\frac{y\cdot 2^x + x \cdot 2^y}{x+y} = 1$$ as we can check that $x + y \neq 0 \Leftrightarrow x \neq \phi, -\frac{1}{\phi}$.

For the case $x > 1 \Rightarrow x > 0 \land y > 0$, we have by AM-GM, there is no solution. I'm not sure how to deal with the case of $x < 1$ i.e. either of $x < 0$ or $y < 0$.