1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Show that ln(x) < sqrt(x) for x>0

  1. Apr 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that ln(x) < √(x) for x>0

    3. The attempt at a solution
    I have no idea where to start! I saw an older thread on here that asked the same question and one suggestion was to take derivatives of both functions and show that the derivative of the RHS would be larger as approaches infinity but how does this show that √(x) is greater for lower values of x?
  2. jcsd
  3. Apr 26, 2013 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    To answer your last question - what does the derivative tell you?
    If that is bigger - what does it mean?
    If it were smaller - what would happen?
  4. Apr 26, 2013 #3


    User Avatar
    Science Advisor
    Homework Helper

    You also need to figure out where the derivative of log(x) starts to be greater than the derivative of sqrt(x). You are going to need a starting point for your argument.
  5. Apr 27, 2013 #4
    The rate of change, so the function with the larger derivative would increase faster...I just wasn't sure using that argument alone would be enough to say that because of that, the function is larger than the other for all x>0.

    I found that the derivatives are equal at x=4 and from there, it looks like the derivative of log(x) is less than the derivtaive of √(x). How would I show that this though? Do I just solve assume
    1/2√(x) > 1/x and solve for x?
  6. Apr 27, 2013 #5
    Hello ChipotleAway,

    Almost there!

    Now that you know that at [itex]x=4[/itex] the derivates are equal, you have two more steps:

    • Prove that at [itex]x=4[/itex], [itex] \sqrt {x} > \ln {x}[/itex]
    • Prove that [itex]\forall x > 4[/itex], [itex] \frac{1}{x} < \frac{1}{2\sqrt {x}}[/itex]

  7. Apr 27, 2013 #6
    Thanks, I'll see if I can show that! Also, wouldn't I need to somehow show that √(x) > log(x) for x between 0 and 4 as well?
  8. Apr 27, 2013 #7
    The case where x is between 0 and 4 is considered trivial.

    Both ##\sqrt{x}## and ##\ln(x)## are monotonically increasing. Because ##\displaystyle \lim_{x\rightarrow 0}\ln(x)<\sqrt{0}## and ##\ln(4)<\sqrt{4}##, we know that ##\ln(x)<\sqrt{x}## from 0 to 4.
  9. Apr 27, 2013 #8


    User Avatar
    2017 Award

    Staff: Mentor

    That is not sufficient. There are functions which satisfy all conditions you mentioned, but violate the inequality to be shown.

    Instead, you can use the derivatives here as well: From 0 to 4, the derivative of ln(x) is larger than the derivative of sqrt(x)....
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted