Proving e^x=x^e for only one positive x

[desquee]

Hi, I'm teaching myself calc 2, and could use some help with a problem I'm not sure how to solve:

Problem:
Prove that x^e=e^x has only one positive solution.

Relevant equations (I think?):
b^x = a^(loga(b)*x) - base change for logarithms

The attempt at a solution:
e^x=x^e=e^(ln(x)*e) - base change
x = ln(x)*e - powers of equal bases
At this point I'm stuck, I'm not sure how to show that x = ln(x)*e has only one positive solution.

 

[Orodruin]

Try to argue for what happens when x < e and when x > e, respectively. (Obviously, x = e is a solution.)

 

[desquee]

I tried assuming x=e+a, and get e^{^e+a}=(e+a)^e=e^ln(e+a)*e, which gives (e+a)/e=ln(e+a), or e^(e+a)/e=e+a, but I haven't been able to reach a clear absurdity.

 

[Orodruin]

Try to use inequalities rather than working with a constant.

 

[Qwertywerty]

Try plotting graph of x∧(1/x) for x>0 .

You will see that it is increasing for x<e and decreasing for x>e .

Now compare x∧(1/x) with e∧(1/e) . The same value is only possible for x=e .

Raise both to x*e and you will get your answer .

 

[Dr. Courtney]

I always start these deals with graphing.

Once you graph it and see that the functions are increasing at different rates, it will become clear how to prove they only intersect once.