i have log_L (1-r)

with L = 2 and r = 0.02

does any1 know how to put log to the base into a calculator or any other way of calculating it?

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- Jan 11th 2007, 03:02 PMquestionanybody good with logs??
i have log_L (1-r)

with L = 2 and r = 0.02

does any1 know how to put log to the base into a calculator or any other way of calculating it? - Jan 11th 2007, 03:37 PMgalactus
You appear to have $\displaystyle log_{2}(1-0.02)=log_{2}(0.98)$

When entering into a calculator, use the change of base formula.

$\displaystyle log_{b}(u)=\frac{log_{a}(u)}{log_{a}(b)}$

In this case, $\displaystyle \frac{log(0.98)}{log(2)}\approx{-0.02914}$

Do not confuse the change of base formula with the quotient rule for logs. - Jan 12th 2007, 07:57 AMquestion
- Jan 12th 2007, 08:11 AMgalactus
No. Where does the 100 come from?.

$\displaystyle log_{2}(1267650600228229401496703205376)=100$ - Jan 12th 2007, 08:42 AMquestion
the answer has to be 100 years its a return period for a flood

- Jan 12th 2007, 09:02 AMCaptainBlack