Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple logarithem

  1. Jul 1, 2011 #1
    if 0 < a < b < 1, is log(a) < log (b) true for all a and b?
    I haven't found any counter example but just to make it sure.
     
  2. jcsd
  3. Jul 1, 2011 #2

    I like Serena

    User Avatar
    Homework Helper

    Hi l-1j-cho! :smile:

    Yes, it is actually true for any 0 < a < b, since the log function is a strictly increasing function.
     
  4. Jul 2, 2011 #3
    thank you!
     
  5. Jul 2, 2011 #4
    take log b - log a and differentiate it.
     
  6. Jul 2, 2011 #5
    It is true for a<b and we know that the domain is (0,infinity) where the base is greater than 1 Because The exponetial function are increasing function with base greater than one and we know that logarithms are inverse functions of exponetial functions try to prove it.

    Thanks
     
  7. Jul 2, 2011 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    It depends upon what you mean by the log function -- what base is being used.

    If 1 < G,
    then log G (x) is strictly increasing on its entire domain, (0, +∞).

    However, 0 < 1/G < 1 and we have log 1/G (x) is strictly decreasing on its entire domain, (0, +∞). You conjuncture would be false in this case​

    So if by the log function, you mean loge a.k.a. ln, or if you mean log10, then the log function is strictly increasing.

    If on the other hand, you are referring to the logarithm function generically, then your conjecture is true if and only if the base is greater than 1.

    It is somewhat unusual to use a logarithm with a base in the range (0, 1).
     
  8. Jul 2, 2011 #7
    I didn't understand what you mean here.Why it is unusual ?
     
  9. Jul 2, 2011 #8

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    It's just unusual to see something like log 0.2 (6), for example.

    If 1 < G, then 0 < 1/G < 1 .

    Doing a change of base gives:
    [tex]\log_{\,1/G}\,(x) =\frac{\log\,_G\,(x)}{\log_{\,\,G}\,(1/G)}=-\log_{\,G}\,(x)[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple logarithem
  1. Simple algegbra? (Replies: 1)

  2. Simple Question (Replies: 4)

  3. A simple question (Replies: 3)

  4. Simple proof (Replies: 8)

Loading...