Proving the Dominance of e^x over x^e: A Mathematical Inquiry

  • Thread starter Echo 6 Sierra
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In summary, our Cal-I Professor asked which is larger, e^x or x^e. It turns out that they are equivalent for all x except e. However, it is possible to find an interval where e^x is larger than x^e.
  • #1
Echo 6 Sierra
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This isn't HW and I don't know what form of math it it but it was posed to us by our Cal-I Professor. First, which is larger?

[tex]e^x[/tex] or [tex]x^e[/tex].

Yes, ^x. But he asks how we as individuals would prove this mathematically? All I can see is to plug in a number for x and go, but I don't think this is what he is after. Would I be hot or cold if I took the log of both and came up with xlne & elnx and then did something, anything else? Either way is just substitution and I'm just rambling now...a tall glass of Port and it's off to bed. :zzz:
 
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  • #2
These functions are equivalent for x = e, so did he define a certain interval for this proof? Not that it would matter in this analytical case, however it is important to know the full problem statement.
 
  • #3
No, he didn't give an interval. He just asked how we would prove one would be greater than the other. I thought it might be something similar to how you would prove/grind out the law of sines, cosines, or a trig identity.

Thank you for the reply. E6S.
 
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  • #4
No, you can't "prove" one is strictly larger than the other because that's not true. As theelectricchild pointed out ex= xe if x= e. However, the graphs are tangent there. it's not too difficult to see that, for all x except e, xe< ex.
 
  • #5
This reminds me of Knoebel's Exponentials Reiterated. I don't recall how Knoebell solves the general question [tex]x^y<y^x[/tex], but by graphing [tex]x^{1/x}[/tex] it's pretty clear that for all [tex]x\neq e[/tex] there are two points where [tex]x^y=y^x[/tex] for fixed x, with one larger inside and the other larger outside the open interval bounded by these two points.
 
  • #6
There is a bit of a question of what your prof meant by larger. If he was just referring to the domain where both are real functions (ie the non-negative reals) then with some playing you get [tex] e^x = x^e \Leftrightarrow x^{1/x}= e^{(e^{-1})} [/tex]. Then you can take the derivative of [tex]x^{1/x} = e^{ln(x)/x}[/tex]. It's not too hard to show that this derivative is positive for x<e and negative for x>e. So you just need values a<e<b such that [tex]e^a > a^e [/tex] and same for b. Since both functions are continuous you are done. You could take a=1 and b=10 since [tex] e^{10} > 2^{10} > 10^3 > 10^e [/tex].

Hope that helps

Steven
 
  • #7
Didn't make it to class last night to get this wrapped up. I'll email him and see if he can tell me how it came out in class.
 

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