Need to prove log x < x for all x > 0, help!!!

[virtuamike]

OK guys, this is stumping me.

log x < x for all x > 0

Don't need to do by mathematical induction, just need to prove. Any ideas?

 

[virtuamike]

No CS majors here?

 

[ThisIsMatt]

dude, I had this last quarter, but I don't remember anything

Try raising both sides by 10

 

[virtuamike]

Hmmm, still can't get it to work

 

[JayHu]

are you sure you have to do it by induction?
i remember doing something similar to this last term (on work term now.. no brain needed..)

 

[virtuamike]

Looked at the question again and I don't need to do proof by induction, just need to prove somehow. Besides induction looks messy, would have to do 3 cases (0 < x < 1, x = 1, x > 1)

 

[BilltheAvenger]

You can prove it by contradiction.

 

[OZEE]

Who cares about proving it by induction... it's that way by definition.

Antilog both... therefore x < 10^x for all x>0.

I dunno... that's been too many years ago. (BSEE '86)

 

[thEnEuRoMancER]

Do it like this:

define f(x) = x - log(x)

prove f(x) > 0

One way to prove this is to calculate the derivative f'(x) and study its behaviour (increasing, falling - I don't know the correct english words).

 

[BruinEd03]

Take the derivative of log x and the derivative of x...for log x it's 1/x and for x it's 1...meaning that as x-> infiniti the slope of log x approaches 0 and slope of x is still 1 therefore we can conclude that x is greater than log x at some finite point in time...which if u just calculate log (1) u can see that it is smaller than x=1 already thus log x < x for all x > 1 hope that helps =)

-Ed

 

[virtuamike]

How would you prove by contradiction? I tried setting it up that way but I can't get it to work.

As far as taking derivatives go, works fine for x > 1. And since log 1 = 0 is less than x = 1, that takes care of case for x = 0. Still leaves 0 < x < 1 (since original problem is for all x > 0).

 

[BruinEd03]

<< How would you prove by contradiction? I tried setting it up that way but I can't get it to work.

As far as taking derivatives go, works fine for x > 1. And since log 1 = 0 is less than x = 1, that takes care of case for x = 0. Still leaves 0 < x < 1 (since original problem is for all x > 0). >>



For 0<x<1 all log x is less than 0 (try it on ur calc)...and since the x is between 0<x<1 log x has to be less than x.

-Ed

 

[thEnEuRoMancER]

Using the derivative, find the minimum of x - log(x) - it's somewhere in interval (0,1) - and calculate the value of x-log(x) in this minimum. It should be greater than zero. Since this is the global minimum, x - log(x) is never smaller than zero and you have your proof.

 

[virtuamike]

Hmmmm, it's been a while since I've done derivatives. To find if a function is increasing or decreasing you just plug and chug into first derivative right? + for increasing and - for decreasing? If that's right then I think I got it.

Also my memory on second derivative is a little blurry. That's to determine concave up or down for a segment, right (+ for up, - for down)?

 

[BruinEd03]

<< Using the derivative, find the minimum of x - log(x) - it's somewhere in interval (0,1) - and calculate the value of x-log(x) in this minimum. It should be greater than zero. Since this is the global minimum, x - log(x) is never smaller than zero and you have your proof. >>



the global minimum is when x approaches 0...if u graph log x it grows exponentially towards negative infiniti as x-> 0 which outstrips x as x-> 0.

virtual mike...I already solved it for you with my two previous posts...i've given u situations for x>=1 and for 0<x<1, which covers x>0 when combined.

-Ed