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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?
  2. jcsd
  3. Jul 16, 2009 #2


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    Staff Emeritus
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    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
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