# Problem with Logarithms solutions

1. Jul 16, 2009

### Jacopo

I know why I can add $$2 \pi i$$ to $$log_{e} \omega$$
($$log_{e} \omega = log_{e}z + i \theta$$ if $$\omega = z(cos \theta +i sen \theta )$$) but I can't understand why I can add $$\frac{2 \pi i}{log_{e}b}$$ to $$log_{b} \omega$$.
Does anyone have the answer for me?
Thanks!

2. Jul 16, 2009

### HallsofIvy

If you know the first, the second is easy! Because
$$log_b(\omega)= \frac{log_e(\omega)}{log_e(b)}= \frac{log_e(\omega)+ i2\pi}{log_e(b)}= \frac{log_e(\omega)}{log_e(b)}+ \frac{2\pi i}{log_e(b)}$$

3. Jul 16, 2009

### Jacopo

Oh yes :D Thanks