logs?

**jason03** · April 22nd 2008, 08:48 AM

HEllo,
Im just trying to remember the algebraic method of sloving for r in an equ. such as this.

a = (1 + (r/m) )^m - 1

**topsquark** · April 22nd 2008, 09:12 AM

Originally Posted by jason03

HEllo,
Im just trying to remember the algebraic method of sloving for r in an equ. such as this.

a = (1 + (r/m) )^m - 1

$a = \left ( 1 + \frac{r}{m} \right )^m - 1$

$a + 1 = \left ( 1 + \frac{r}{m} \right )^m$

$1 + \frac{r}{m} = (a + 1)^{1/m}$

$\frac{r}{m} = (a + 1)^{1/m} - 1$

$r = \frac{1}{m} \left ( (a + 1)^{1/m} - 1 \right )$

-Dan

**jason03** · April 22nd 2008, 10:11 AM

Hello,
I tried your equ. but im not coming up with the correct answer.

heres the equ with the numbers

.0931 = (1 + r/4)^4 - 1

the answer is .09 and it works when plugged into the above equ.

But I tried using your equ. and I get a different answer

**Moo** · April 22nd 2008, 10:16 AM

Hello,

Originally Posted by topsquark

$r = \frac{1}{m} \left ( (a + 1)^{1/m} - 1 \right )$

-Dan

There is just a small mistake here.

It's $r={\color{red}m} \left ( (a + 1)^{1/m} - 1 \right )$

**jason03** · April 22nd 2008, 07:59 PM

ohh Thank you!

If you dont mind me being a pest, how did you arive at that. I follow you up until the exponents change.

**Moo** · April 23rd 2008, 12:27 AM

Originally Posted by topsquark

$a = \left ( 1 + \frac{r}{m} \right )^m - 1$

$a + 1 = \left ( 1 + \frac{r}{m} \right )^m$

power everything to 1/m. As $(a^b)^c=a^{bc}$ , $((\dots)^m)^{1/m}=(\dots)$

$1 + \frac{r}{m} = (a + 1)^{1/m}$

$\frac{r}{m} = (a + 1)^{1/m} - 1$

multiply by m in each side

$r = {\color{red} m} \left ( (a + 1)^{1/m} - 1 \right )$

-Dan

Is it ok ?

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