Find $\alpha>0$ such that $e^x>x^\alpha$ $\forall x>0$

Question

How find which $\alpha>0$ satisfies $e^x>x^\alpha$ $\forall x>0$?

I know that $$e^x=\left(e^\frac{x}{\alpha}\right)^\alpha\geq \left(1+\frac{x}{\alpha}\right)^\alpha\geq \frac{x^ \alpha}{\alpha^\alpha}.$$ The RHS is surely greater or equal to $x^\alpha$ if $\alpha<1$. But how can I study the remaining cases?

user · Accepted Answer · 2024-02-06T16:08:58.950

2

We have that

$$f(x)=e^x-x^e=e^x-e^{e\log x}>0 \iff g(x)=x-e\log x>0 $$

and $g(x)\ge 0$ with $g(x)=0 \iff x=e$ therefore, since for $\alpha=e$ we have exactly one root, the solution is $\alpha<e$.

edited Feb 06 '24 at 16:08

answered Feb 06 '24 at 15:20

user

162,563

Okk but this is $\alpha=e$, I want to study all possible $\alpha>0$, so proving where the inequality works or not – axi Feb 06 '24 at 15:21
@axi For $\alpha=e$ we have exactly one root then the solution is $\alpha>e$. – user Feb 06 '24 at 15:23
why if $\alpha=e$ give me one solution then I have to take all $\alpha>e$? – axi Feb 06 '24 at 15:38
Yes the problem is equivalent to $x-\alpha \log x>0$. – user Feb 06 '24 at 15:40
So if $\alpha>e$ the inequality holds not for all x>0 as required... – axi Feb 06 '24 at 15:44
The solution is $\alpha < e$. – Jan Feb 06 '24 at 16:07
1

@Jan Yes of course! I reversed the inequality in my mind! Thanks – user Feb 06 '24 at 16:08
@user okk thanks...the only ting that I dcan't understand is: why we select $\alpha=e$, really in general for $x=\alpha$ then $g(x)>0$, so why in particular $x=\alpha=e$? – axi Feb 06 '24 at 19:20
For symmetry we see that this leads to $0$ then it’s a good guess. But we can also proceed differently. – user Feb 06 '24 at 19:25
okk many thanks – axi Feb 06 '24 at 19:30

TonyK · Answer 2 · 2024-02-06T16:19:56.103

2

$x^\alpha=e^{\alpha\log x}$. And the function $e^x$ is monotonic increasing. So $e^x>x^\alpha\iff x>\alpha\log x$.

We want to find all $\alpha>0$ for which $x>\alpha\log x$ whenever $x>0$. So define $f_\alpha(x)=x-\alpha\log x$. This function tends to $+\infty$ as $x\downarrow 0$ or $x\to+\infty$. So we want to find its minimum, which occurs when $f'_\alpha(x)=1-\alpha/x=0$, i.e. when $x=\alpha$. And at this minimum, we have $$f_\alpha(x)=f_\alpha(\alpha)=\alpha(1-\log\alpha)$$ You can take it from there.

edited Feb 06 '24 at 16:19

answered Feb 06 '24 at 16:08

TonyK

68,059

Ok so when I have $f_\alpha<f$ then how can I conclude that the inequality is satified $\forall \alpha<e$? – axi Feb 06 '24 at 17:18
I mean how can I arrive to take from $x=\alpha$, then $\alpha=e$!? – axi Feb 06 '24 at 17:42

score 0 · Answer 3 · answered Feb 06 '24 at 16:35

Given is $e^x>x^\alpha,\forall x\ge0$, taking logarithm on both sides, we get $x>\alpha \log{x}$. Now if $0<x<1$, $\log{x}<0$, so if we divide by $\log{x}$ the inequality gets reversed: $\frac{x}{\log{x}} < \alpha$.

Thus, the value of $\alpha$ must be greater than the maximum value of $\frac{x}{\log{x}}$ in $(0,1)$, which is at $x=0$ and the maximum value in $(0,1)$ is $0$, as we can see from this graph. Thus $\alpha > 0$.

Now for the case when $x>0$, $\log{x}>0$, thus $\frac{x}{{\log}{x}}>\alpha$. In this case, $\alpha$ must be smaller than the minimum value of $\frac{x}{\log{x}}$ in $(1,\infty)$, which is at $x=e$ and the value of the minimum is $e$. Thus, $\alpha<e$.

Overall, we get that $0<\alpha<e$, such that $e^x>x^\alpha,\forall x>0$.

Find $\alpha>0$ such that $e^x>x^\alpha$ $\forall x>0$

3 Answers3