I'm trying to find the interval where $e^x$ grows faster than $x^x$. So, i need to solve the inequality: $$\frac{d}{dx}e^x > \frac{d}{dx}x^x$$ Taking the derivatives, i arrived to this inequality: $$e^x > x^x(\operatorname{ln}(x) + 1)$$ And i don't know how to start here. I've tried going this way: $$x^x = e^{x \operatorname{ln}(x)}$$ $$e^x > e^{x \operatorname{ln}(x)} (\operatorname{ln}(x) + 1)$$ $$e^{x-x\operatorname{ln}(x)} > \operatorname{ln}(x)+1$$ $${(e^{1-\operatorname{ln}(x)})}^x> \operatorname{ln}(x) + 1$$ $$\bigg(\frac{e}{x}\bigg)^x > \operatorname{ln}(x) + 1$$ And now i don't know where to go. Did i started correctly? If i did, where should i go now? If i didn't, what is the best way to approach this inequality? Thanks.

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    $\begingroup$ but you have an intersection Point of These two graphs $\endgroup$ – Dr. Sonnhard Graubner Mar 19 '17 at 20:12
  • $\begingroup$ Looks like $0<x<2.08711$ numerically, but that's far from a closed form... $\endgroup$ – coffeemath Mar 19 '17 at 20:17
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    $\begingroup$ Your last two inequalities are wrong. $\endgroup$ – Dr. Wolfgang Hintze Mar 19 '17 at 20:27
  • $\begingroup$ Dr. Wolfgang Hintze, true. I'll correct them soon. $\endgroup$ – embedded_dev Mar 19 '17 at 22:30
  • $\begingroup$ I believe they are correct now. $\endgroup$ – embedded_dev Mar 19 '17 at 23:03

For $x\in (0,1/e]$ we have $1+\ln x\leq 0$ and $x^x>0$ so $e^x>0\geq x^x(1+\ln x).$

For $x\in (1/e,1]$ we have $0<1+\ln x\leq 1$ and $x^x\leq 1$ and $e^x>1$ so $e^x>1\geq x^x(1+\ln x).$

For $x\geq 1$ let $g(x)= \frac {e^x}{x^x(1+\ln x)}.$ We have $$(\ln g(x))'=-\ln x-\frac {1}{x(1+\ln x)}<0$$ so $\ln g(x)$ is strictly decreasing on $[1,\infty)$. Therefore $g(x)$ is strictly decreasing on $[1,\infty),$ with $g(1)=e>1$ and $g(e)= 1/2<1.$ So there is a unique $x_0\in (1,e)$ with $g(x_0)=1$. Since $g$ is strictly decreasing on $[1,\infty)$ we have $$x\in [1,x_0)\implies g(x)>g(x_0)=1,$$ $$\text {and }\quad x\geq x_0\implies g(x)\leq g(x_0)=1.$$

In summary, $e^x>x^x(1+\ln x)\iff x\in (0,x_0).$

Note: $g(x_0)=1\iff \ln g(x_0)=0\iff x_0-x_0\ln x_0-\ln (1+\ln x_0)=0.$ A quick calculation shows that $x_0>2. $

  • $\begingroup$ Thank you, it is a very interesting approach. A question: $x_0-x_0\ln x_0-\ln (1+\ln x_0)=0$ will give me the exact value $2.08711$, right? If so, how can i solve that equation? $\endgroup$ – embedded_dev Mar 20 '17 at 14:17
  • $\begingroup$ Right. Since $\ln g(x)$ is strictly decreasing on $[1,\infty),$ Newton's Method will give good numerical approximations. I don't think It's the kind of equation with a closed-formula solution in terms of elementary functions. $\endgroup$ – DanielWainfleet Mar 20 '17 at 19:55
  • $\begingroup$ That's what i was thinking, no "non-numerical" way of solving this. Thanks. $\endgroup$ – embedded_dev Mar 20 '17 at 22:38

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