9
$\begingroup$

How to solve this equation $$x^{2}=2^{x}$$

where $x \in \mathbb{R}$.

Por tentativa erro consegui descobri que $2$ é uma solução, mas não encontrei um método pra isso. Alguma sugestão?(*)

(Translation: By trying different values I've found that $2$ is a solution, but I couldn't find any method to this though. Any suggestions? )

$\endgroup$
4
  • 6
    $\begingroup$ $2,4$ are trivial, $-0.76666469596212309311\dots$ is obtained either by the Lambert W function or approximated by Newton method. $\endgroup$
    – UserX
    Oct 9, 2014 at 22:02
  • 1
    $\begingroup$ This problem plagued me for years, finding the third root. Only learned about the lambert omega function 2 years ago :) $\endgroup$
    – Alan
    Oct 9, 2014 at 22:05
  • 2
    $\begingroup$ يعني عادي الواحد يتكلم اي لغة هنا؟ $\endgroup$
    – seteropere
    Oct 9, 2014 at 23:24
  • $\begingroup$ The same question is treated here: math.stackexchange.com/questions/591124/solve-2x-x2 $\endgroup$ Mar 23, 2016 at 23:17

4 Answers 4

18
$\begingroup$

The equation can be written $x\log2=2\log|x|$. Let's consider the function $$ f(x)=x\log2-2\log|x| $$ defined for $x\ne0$. We have easily $$ \lim_{x\to-\infty}f(x)=-\infty, \qquad \lim_{x\to\infty}f(x)=\infty $$ and $$ \lim_{x\to0}f(x)=\infty. $$ Moreover $$ f'(x)=\log2-\frac{2}{x}=\frac{x\log2-2}{x} $$ Set $\alpha=2/\log2$; then $f'(x)$ is positive for $x<0$ and for $x>\alpha$, while it's negative for $0<x<\alpha$.

Thus the function is increasing in $(-\infty,0)$, which accounts for a solution in this interval. In the interval $(0,\infty)$ the function has a minimum at $\alpha$ and $$ f(\alpha)=\frac{2}{\log2}\log2-2\log\frac{2}{\log2} =2(1-\log2+\log\log2)\approx-0.85 $$ Since the minimum is negative, this accounts for two solutions in $(0,\infty)$, which clearly are $x=2$ and $x=4$.

$\endgroup$
9
$\begingroup$

Try this, first suppose $x > 0$, then you take $x<0$.

$$\begin{align}x^2 = 2^x &\Rightarrow (x^2)^{\frac{1}{2}} = (2^x)^\frac{1}{2} \Rightarrow x= 2^\frac{x}{2} \\ & \Rightarrow x \ e^{-x\frac{\ln\ 2}{2}} = 1 \Rightarrow -x \frac{\ln\ 2}{2}\ e^{-x\frac{\ln\ 2}{2}} = -\frac{\ln \ 2}{2} \\ &\Rightarrow -x \frac{\ln\ 2}{2} = W(-\frac{\ln\ 2}{2}) \Rightarrow x = -\frac{2\ W(-\frac{\ln \ 2}{2})}{\ln\ 2}\end{align}$$

Which gives us

$x = -\frac{2\ W(\frac{-\ln \ 2}{2})}{\ln\ 2} = 2$ , in case $x > 0$

Similarly we may find

$x = -\frac{2\ W(\frac{\ln \ 2}{2})}{\ln\ 2} \approx -0,76666$, in case $x < 0$

Where $W$ is the Lambert's funtion.

$\endgroup$
6
  • $\begingroup$ You might consider explaining how you obtained all three roots from this approach. It seems that one only obtains 1 root for $x<0$ and one other for $x>0$. Also, $-\frac{2W(-log(2)/2)}{\log(2)}=2$. You have a sign error and the incorrect root I believe. $\endgroup$
    – Mark Viola
    Jan 4, 2016 at 17:52
  • $\begingroup$ @Dr.MV I adressed to your observations. Thank you for pointing it out. $\endgroup$ Jan 4, 2016 at 18:46
  • $\begingroup$ @pitagoras When you posted this, did n't you get an indication of the same question in the past? $\endgroup$
    – Narasimham
    Jan 4, 2016 at 18:48
  • $\begingroup$ @Narasimham No, unfortunately. $\endgroup$ Jan 4, 2016 at 18:49
  • $\begingroup$ You're welcome. My pleasure. There is one more, of course, one more root, $x=4$. How do you account for it using this approach? $\endgroup$
    – Mark Viola
    Jan 5, 2016 at 3:02
0
$\begingroup$

I wrote about this in another post.

The other solution is 4.

For $x^a = a^x; a > 0$ in general... Well, notice the shapes of the graphs for $x \ge 0$ are such that they intersect twice if $a \ne e$ and intersect once if $a = 1$ (at $x = e$).

In my other post I went into great detail about taking the first and second derivatives to show both $a^x$ and $x^a$ are concave and that they can only have at most two intersections and as $a^0 = 1 > 0^a = 0$ there must have at least one. But I'll leave that as an excercise to the reader this time.

But obviously $x = a \implies a^a = a^a$ is a solution. The other ... hmm... I forget and I'm too lazy to figure it out a second time. But the solutions are "paired" (that is if $x = b$ solves $a^x = x^a$ then $x = a$ solves $b^x = x^b$) and one solution is less than e and the other is greater than e. I came up with a formula relating the solutions.

2 and 4 solves both $x^2 = 2^x$ and $x^4 = 4^x$ were the slickest solutions but every other positive value for $a$ and a $a$ and $a'$ solution.

If $a$ is even or an even rational then there is an $x = -1/a$ solution. But if a is odd then there is no negative solution as for $x < 0$ then $a^x > 0$ but $x^a < 0$. If a is irrational then $x^a$ is undefined for $x < 0$. These are the only solutions as $a^x$ is increasing but $x^a$ is decreasing when $x < 0$ and $a$ is even.

$\endgroup$
0
$\begingroup$
We see by observation that x = 2 and x = 4 are clearly solutions. 
We will prove by induction that 2^n > n^2 for all n >= 5. 
For the base case, n = 5 gives 32 > 25 which is true.
For the inductive case, we know that 2^n > n^2 and we want to show that 2^(n+1) > (n+1)^2. 
We will have that 2^(n+1) > 2*n^2 and because 2*n^2 = n^2 + n^2 > n^2 + 2*n + 1 when n >=5
(n^2 - 2*n - 1 = (n-1)^2 - 2 > 0 for n >= 5) we will also have that 2^(n+1) > (n+1)^2 as we wanted 
to show. By induction 2^n > n^2  for n >= 5 and the only solutions are x=2 and x = 4.
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .