i have

r = [1 - ( 1 - 1/Tr)^L ] x 100%

and i need Tr and i have L = 2 and r = 2%

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- Jan 13th 2007, 09:16 AMquestionhelp lease
i have

r = [1 - ( 1 - 1/Tr)^L ] x 100%

and i need Tr and i have L = 2 and r = 2% - Jan 13th 2007, 09:51 AMCaptainBlack
- Jan 13th 2007, 10:02 AMtopsquark
$\displaystyle r = [1 - ( 1 - 1/Tr)^L ] \cdot 100$

$\displaystyle \frac{r}{100} = 1 - \left ( 1 - \frac{1}{Tr} \right )^L$

$\displaystyle \frac{r}{100} - 1 = - \left ( 1 - \frac{1}{Tr} \right )^L$

$\displaystyle 1 - \frac{r}{100} = \left ( 1 - \frac{1}{Tr} \right )^L$

(Just a note: I seem to recall doing this before and doing something silly with a "log" term at this point. I don't know why I did it that way, and it may well have been wrong. My apologies if I'm remembering that correctly!)

$\displaystyle \left ( 1 - \frac{r}{100} \right ) ^{1/L} = 1 - \frac{1}{Tr}$

$\displaystyle \frac{1}{Tr} = 1 - \left ( 1 - \frac{r}{100} \right ) ^{1/L}$

$\displaystyle Tr = \frac{1}{1 - \left ( 1 - \frac{r}{100} \right ) ^{1/L}}$

With r = 2 and L = 2:

$\displaystyle Tr = \frac{1}{1 - \left ( 1 - \frac{2}{100} \right ) ^{1/2}}$

$\displaystyle Tr = \frac{1}{1 - \sqrt{\frac{98}{100}}}$

$\displaystyle Tr = 99.4975$

-Dan