To somehow prove that x>logx for all x>o

[khurram usman]

i have been given a question ...that is to prove that x>logx for all x>0
i know that its true...and i have thought of a way to prove it
i thought of drawing a graph...but then i had a small doubt that will the graph hold true for all x>0 till infinity?
i mean i know its true but will the graph prove that? i can only draw it for a finite no of numbers
also i am studying differentiation and integration these days...so is thereany way of roving this using differentiation and integration? ...or any other way?
i really appreciate anyones help...thanks

 

[Pi-Bond]

In order to prove it, you can proceed by showing x-log(x)>0 for all x>0, and then showing that x-log(x) always increases (by using differentiation; it must always have a positive slope).

 

[khurram usman]

In order to prove it, you can proceed by showing x-log(x)>0 for all x>0, and then showing that x-log(x) always increases (by using differentiation; it must always have a positive slope).

but then again i will be doing that for a finte limit...
i want to do something with limits like x goes from 0 to infinity...any other idea?

 

[Pi-Bond]

Taking a limit at infinity proves nothing in this case; it will only give you information for large x behavior, and nothing about the general behavior. What I told you applies on the whole interval.

 

[Trevor Vadas]

You know that x>lnx for 0<x<=1. Try using derivatives to show the function x increases faster than ln(x) for all x >1.

 

[flyingpig]

induction...?

 

[Pi-Bond]

induction...?

I don't think induction can be used for statements such as the OP's (the functions are continuous; they aren't just defined for natural numbers)

 

[snipez90]

You could also divide by x (since x > 0) and look at the behavior of log(x)/x.