Which increases faster e^x or x^e ?

1. Mar 26, 2012

Roni1985

1. The problem statement, all variables and given/known data
which increases faster e^x or x^e ?

2. Relevant equations

3. The attempt at a solution

My attempt was taking the log of both, assuming it doesnt change anything (is this assumption correct?)

x*ln(e) ------------------------ e*ln(x)

now I took the derivative

1 ------------------------ e/x

and I said that if
0<x<e, x^e increases faster
otherwise, e^x is faster

Is my logic correct?

Thanks

2. Mar 26, 2012

Vorde

No, you can't just take the log of both. You aren't dealing with an equality you are comparing two equations. For specific values, your procedure of deriving both and comparing when one is greater works fine, but you have to derive the original equations.

In general though, exponential growth (blank^x) grows faster than x^blank (whatever that is called), and you can prove that with limits.

3. Mar 26, 2012

Roni1985

How can you prove it with limits?
lim x-> inf
and then lim x-> -inf
?

4. Mar 26, 2012

Gengar

Or simply consider derivatives:
d/dx (e^x) = e^x and d/dx(x^e)=ex^(e-1)
These tell you how fast each function is increasing at each x, hence you can work out which function is increasing faster at each x.

5. Mar 26, 2012

Roni1985

But you are using the fact:

"exponential growth (blank^x) grows faster than x^blank (whatever that is called)"

I want to prove it.
how do you see that
e^x grows faster then ex^(e-1) ?

6. Mar 26, 2012

Vorde

You have to consider the limit of the ratio of the two equations as x-->inf. Using l'hopital's rule, you can see the difference in the rates of change.