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!

Problem with Logarithms solutions

  1. Jul 16, 2009 #1
    I know why I can add [tex]2 \pi i[/tex] to [tex] log_{e} \omega [/tex]
    ([tex] log_{e} \omega = log_{e}z + i \theta [/tex] if [tex] \omega = z(cos \theta +i sen \theta ) [/tex]) but I can't understand why I can add [tex] \frac{2 \pi i}{log_{e}b} [/tex] to [tex] log_{b} \omega [/tex].
    Does anyone have the answer for me?
    Thanks!
     
  2. jcsd
  3. Jul 16, 2009 #2

    HallsofIvy

    User Avatar
    Science Advisor

    If you know the first, the second is easy! Because
    [tex]log_b(\omega)= \frac{log_e(\omega)}{log_e(b)}= \frac{log_e(\omega)+ i2\pi}{log_e(b)}= \frac{log_e(\omega)}{log_e(b)}+ \frac{2\pi i}{log_e(b)}[/tex]
     
  4. Jul 16, 2009 #3
    Oh yes :D Thanks
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook