Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Total no. of real solution of the equation $x^2 = 2^{-x}$

My Try:: Let $f(x) = x^2-2^{-x}$

Diff. both side $x\;,$ we get

$f^\prime(x) = 2x+2^{-x}\ln(2)$

$f^{\prime\prime}(x)=2+2^x.\ln(2)>0 \; \forall x\in \mathbb{R}$

Here $f^{\prime\prime}(x)>0\forall x\in \mathbb{R}$

So $f(x)$ is an Concave upward function that means if function $f(x)$ has real roots then it must be $2$

But using wolframalpha We get $3$ real solution.

So were is my mistake


share|cite|improve this question
As I told you on your previous question: you're supposed to accept your favorite answers, if they're satisfactory. You've accepted $0$ answers on $13$ questions so far. – Git Gud Jan 26 '13 at 11:07
Oh Sorry Git Gud – juantheron Jan 26 '13 at 11:21
Don't apologize to me. Go to your profile and accept answers for your previous questions. – Git Gud Jan 26 '13 at 11:27
How can i accept answer would you like to explain me .i was thinking that votes means accepting answer. – juantheron Jan 26 '13 at 11:39
No. When you over your mousse below the downvote arrow, a 'check' sign will appear. You should click on it to accept an answer. If you do this properly, the check sign will be visible permanently and it will turn green. – Git Gud Jan 26 '13 at 11:40
up vote 1 down vote accepted

Another approach is using the Lambert-W function.

Since $x^2=2^{-x}\Rightarrow \log(2)x/2\,e^{\log(2)x/2}=\pm\log(2)/2$, we get that $$ x=\frac2{\log(2)}\mathrm{W}\left(\pm\frac{\log(2)}{2}\right) $$ where $\mathrm{W}$ is the Lambert-W function.

For positive arguments, $\mathrm{W}$ only has one real branch, so the only positive real root of the equation is N[2/Log[2] LambertW[0, Log[2]/2], 20]: $$ x\doteq0.76666469596212309311 $$ For negative arguments, $\mathrm{W}$ has two real branches. The negative real solutions are N[2/Log[2] LambertW[0, -Log[2]/2], 20] $$ x=-2 $$ and N[2/Log[2] LambertW[-1, -Log[2]/2], 20] $$ x=-4 $$

share|cite|improve this answer

Because $$ f''(x)=2-2^{-x}\ln^2 2 $$ and it is not always positive. You need differentiate one more time to see that $f'''>0$ and conclude that $f$ have at most three solutions

share|cite|improve this answer
Thanks Norbert. $f''(x)=2-2^{-x}\ln^2(2)$ and $f'''(x)=2^{-x}\ln^2(3)>0$ means $f'(x)$ is Concave upward function So It has at most $2$ solution.So how can i prove that $f(x)$ has at most $3$ solution. – juantheron Jan 26 '13 at 11:25
Suppose $f$ has four roots or more. Take a set of four consecutive roots in increasing order $\{a,b,c,d\}$. What does Rolle tell you about $f(a)$ & $f(b)$? What about $f(b)$ & $f(c)$? And finally, what about $f(c)$ & $f(d)$? – Git Gud Jan 26 '13 at 11:30

Let $f(x)=x^2-2^{-x}$, then $f''(x)=2-(\log2)^22^{-x}$. Since $f''$ is increasing from $f''(-\infty)=-\infty$ to $f''(+\infty)=+\infty$, $f'$ is decreasing on $(-\infty,x]$ and increasing on $[x,+\infty)$, for some $x$. Two cases can occur.

  • Either $f'(x)\geqslant0$, then $f'$ is nonnegative everywhere and $f$ is increasing. Since $f(-\infty)=-\infty$ and $f(+\infty)=+\infty$, this implies that $f=0$ has exactly one root.

  • Or $f'(x)\lt0$, then $f'$ is positive then negative then positive.

Since $f(-3)\gt0$ and $f(0)\lt0$, the second case occurs. Thus, the function $f$ is negative then positive then negative then positive, and it has exactly three zeroes, one in each interval $(-\infty,-3)$, $(-3,0)$ and $(0,+\infty)$. The zeroes in the first two intervals are easy-to-guess integers, and the positive zero happens to lie in the interval $(\frac12,1)$ since $f(\frac12)\lt0\lt f(1)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.