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

  • Thread starter khurram usman
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In summary, the conversation discusses different ways to prove that x>log(x) for all x>0. The first suggestion is to show that x-log(x)>0 and that it always increases, using differentiation. However, it is noted that this approach only works for a finite limit and may not demonstrate the general behavior. Another idea is to use limits at infinity, but this is also deemed insufficient. The suggestion to use induction is dismissed as it cannot be used for continuous functions. Another approach is to consider the behavior of log(x)/x by dividing by x and examining its behavior.
  • #1
khurram usman
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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
 
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  • #2
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).
 
  • #3
Pi-Bond said:
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?
 
  • #4
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.
 
  • #5
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.
 
  • #6
induction...?
 
  • #7
flyingpig said:
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)
 
  • #8
You could also divide by x (since x > 0) and look at the behavior of log(x)/x.
 

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

How can we prove that x>logx for all x>0?

To prove this statement, we can use mathematical induction. First, we will show that it is true for x=1. Then, we assume that it is true for some arbitrary value of x, and use this assumption to prove that it is also true for x+1. This will show that the statement is true for all positive integers, and since x can take on any positive value, it will also be true for all x>0.

Can we use a counterexample to disprove this statement?

No, because the statement is true for all positive values of x. A counterexample would require a specific value of x that would make the statement false, but since x can take on any positive value, there is no single value that would disprove the statement.

What is the relationship between x and logx?

The relationship between x and logx is that as x increases, logx increases at a slower rate. This means that for any value of x>0, x will always be greater than logx.

How does the graph of x and logx illustrate this statement?

The graph of x and logx will show that as x increases, logx also increases, but at a slower rate. This is because the logarithmic function is an inverse function of an exponential function, and as x increases, the exponential function will grow at a faster rate than the logarithmic function.

Can we use any other methods to prove this statement?

Yes, we can also use calculus to prove this statement. By taking the derivative of both x and logx, we can show that the derivative of x is always greater than the derivative of logx. This means that x will always be increasing at a faster rate than logx, and therefore x will always be greater than logx for all x>0.

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