Which increases faster e^x or x^e ?

[Roni1985]

Homework Statement

which increases faster e^x or x^e ?

Homework Equations

The Attempt at a Solution

My attempt was taking the log of both, assuming it doesn't 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

 

[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.

 

[Roni1985]

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.

Thanks for the reply.

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

 

[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.

 

[Roni1985]

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.

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) ?

 

[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.