Number of real solution of $e^x=x^n\;,$ Where $n\in \mathbb{N}$

$\bf{My\; Try::}$ If $n=1\;,$ Then we get $e^x=x$

Now Let $f(x)=e^x-x\;,$ Then $f'(x) = e^x-1\;,$ and $f''(x)=e^x>0\;\forall\; x\in \mathbb{N}$

So $f''(x)=0$ has no real Roots, Then Using $\bf{LMVT}$, We get $f'(x)=0$ has at most one real roots

And $f(x)=0$ has at most $2$ real roots.

But Using Wolframalpha, We get no real roots.(I did not understand how can i calculate it.)

Now Put $n=2\;,$ We get $e^x=x^2$

Now let $f(x)=e^x-x^2\;,$ Then $f'(x)=e^x-2x$ and $f''(x)=e^x-2$ and $f'''(x)=e^x>0\;\forall\; x\in \mathbb{R}$

So $f'''(x)=0$ has no real roots And again Using $\bf{LMVT}\;,$ We get $f''(x)=0$

has at most one real roots and $f'(x)=0$ has at most $2$ real roots and $f(x)=0$

has at most $3$ real roots

Now Here $f(x)=e^x-x^2$ has one root $x\in (0,1)$ Now i did not understand How can I calculate

In which interval other roots Lie

And How can I calculate no. of real Roots of $e^x=x^n$ for $n\in \mathbb{N}$

Help required, Thanks

  • 1
    $\begingroup$ Note that for n=1, there is no intersection between the graph of $e^x$ and that of the first bisector $\endgroup$
    – A s
    May 17 '16 at 11:16
  • 1
    $\begingroup$ This may help: math.stackexchange.com/questions/646092/… $\endgroup$ May 17 '16 at 11:17
  • 1
    $\begingroup$ For even $n\gt e$ you might reasonably expect three solutions, one negative and two positive. For odd $n\gt e$ you might reasonably expect two solutions, both positive. $\endgroup$
    – Henry
    May 17 '16 at 11:18
  • $\begingroup$ Consider the function (e^x)/(x^n) : its derivative, and for what x it is equal to 1 ? $\endgroup$
    – charmd
    May 17 '16 at 11:21
  • $\begingroup$ Use the Lambert $W$ function to solve it in terms of $n$, which will then allow you to determine any real solutions (if any). $\endgroup$ May 17 '16 at 11:21

The derivate of $$f(x)=e^x-x^n$$

is $$f'(x)=e^x-nx^{n-1}$$

If $n$ is even, we have $f'(x)>0$ for all $x\le 0$, so there is at most one negative solution.

Because of $f(0)=1$ and $f(-1)=\frac{1}{e}-1<0$ for even $n$, there is exactly one negative root for even $n$.

If $n$ is odd, there is no negative root because of $f(x)>0$ for all $x\in\mathbb R$

For positive $x$, we can use the equation $g(x)=x-n\cdot ln(x)=0$

The first derivate is $1-\frac{n}{x}$, the second derivate is $\frac{n}{x^2}$, hence always positive.

For $n=1$ and $n=2$, the minimum of $g(x)$ , which is at $x=n$, is positive, so there is no positive root. If $n\ge 3$, then $n-n\log(n)$ is negative, so there are two positive roots.


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