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

[Echo 6 Sierra]

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?

$$e^x$$ or $$x^e$$.

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:

 

[Theelectricchild]

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.

 

[Echo 6 Sierra]

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.

 

[HallsofIvy]

No, you can't "prove" one is strictly larger than the other because that's not true. As theelectricchild pointed out e^x= x^e if x= e. However, the graphs are tangent there. it's not too difficult to see that, for all x except e, x^e< e^x.

 

[CRGreathouse]

This reminds me of Knoebel's Exponentials Reiterated. I don't recall how Knoebell solves the general question $$x^y<y^x$$, but by graphing $$x^{1/x}$$ it's pretty clear that for all $$x\neq e$$ there are two points where $$x^y=y^x$$ for fixed x, with one larger inside and the other larger outside the open interval bounded by these two points.

 

[snoble]

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 $$e^x = x^e \Leftrightarrow x^{1/x}= e^{(e^{-1})}$$. Then you can take the derivative of $$x^{1/x} = e^{ln(x)/x}$$. 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 $$e^a > a^e$$ and same for b. Since both functions are continuous you are done. You could take a=1 and b=10 since $$e^{10} > 2^{10} > 10^3 > 10^e$$.

Hope that helps

Steven

 

[Echo 6 Sierra]

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.