# anybody good with logs??

• January 11th 2007, 03:02 PM
question
anybody 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?
• January 11th 2007, 03:37 PM
galactus
You appear to have $log_{2}(1-0.02)=log_{2}(0.98)$

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

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

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

Do not confuse the change of base formula with the quotient rule for logs.
• January 12th 2007, 07:57 AM
question
Quote:

Originally Posted by galactus
You appear to have $log_{2}(1-0.02)=log_{2}(0.98)$

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

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

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

Do not confuse the change of base formula with the quotient rule for logs.

having Tr = 1/1-log_L(1-r)

with L = 2 and r = 0.02

i get this to be equal to 0.97168 and i know the answer is approx 100 so should my answer be multiplied by 100?? is this the rite procedure?
• January 12th 2007, 08:11 AM
galactus
No. Where does the 100 come from?.

$log_{2}(1267650600228229401496703205376)=100$
• January 12th 2007, 08:42 AM
question
the answer has to be 100 years its a return period for a flood
• January 12th 2007, 09:02 AM
CaptainBlack
Quote:

Originally Posted by question
the answer has to be 100 years its a return period for a flood

Perhaps you should post the actual question you are trying to answer.

RonL