1. ## logs?

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

2. 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
$\displaystyle a = \left ( 1 + \frac{r}{m} \right )^m - 1$

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

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

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

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

-Dan

3. 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

4. Hello,

Originally Posted by topsquark
$\displaystyle r = \frac{1}{m} \left ( (a + 1)^{1/m} - 1 \right )$

-Dan
There is just a small mistake here.

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

5. 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.

6. Originally Posted by topsquark
$\displaystyle a = \left ( 1 + \frac{r}{m} \right )^m - 1$

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

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

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

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

multiply by m in each side

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

-Dan
Is it ok ?