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

Convergence in logarithms

  1. Sep 24, 2010 #1

    If I have ln{a} converging to ln{b}, can I claim that a converges to b?

    I would think yes, because ln is a monotonic function and we are always sure to get a unique value of ln for any argument 'a'. Is that reasoning correct?

    I am not able to prove it by the definition of convergence,

    By convergence in logarithm we get: ln(a) - ln(b) < \epsilon,

    ln{a/b} < \epsilon

    a/b < exp{\epsilon}

    (a-b)/b < exp{\epsilon} - 1

    a-b < b(exp{\epsilon} - 1)

    From here I can claim that since \epsilon is near to zero exp{\epsilon} is near to 1 and hence exp{\epsilon} - 1 < \epsilon_1


    a - b <b \epsilon_1.

    Can I claim from here and a converges to b? The presence of b on the right hand side bothers me. Any help will be greatly appreciated.

    Thanks a lot...
  2. jcsd
  3. Sep 24, 2010 #2


    User Avatar
    Science Advisor

    You should work with absolute values. As far as the multiplication by b, it doesn't affect the analysis as long as |b| > 0.
  4. Sep 25, 2010 #3
    Thanks for the reply. I forgot to mention that a,b >0. So, as long as b is bounded, the analysis should work, right?
  5. Sep 25, 2010 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If you want to use a slightly "higher level" argument:

    If ln(an) converges to ln(a), then since exponentiation is continuous, [tex]e^{ln(a_n)}[/tex] converges to [tex] e^{ln(a)}[/tex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook