I would like to understand how to solve for r in this problem.

2773=1000(1+r)^36

I know that r=.0287 but I do not understand how to solve with the exponents. Do I use Ln?(Headbang) and how is that done? Thanks

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- Sep 13th 2008, 12:36 PMSmoriginalWhat is the math behind this?
I would like to understand how to solve for r in this problem.

2773=1000(1+r)^36

I know that r=.0287 but I do not understand how to solve with the exponents. Do I use Ln?(Headbang) and how is that done? Thanks - Sep 13th 2008, 12:47 PMShyam
yes you have to use ln, see here,

$\displaystyle

\frac{2773}{1000}=(1+r)^{36}$

$\displaystyle 2.773=(1+r)^{36}$

Now take natural log on both sides:

$\displaystyle

\ln{2.773}= \ln \left[(1+r)^{36}\right]$

$\displaystyle

1.019929767= 36 \ln (1+r)$

$\displaystyle

\frac{1.019929767}{36}= \ln (1+r)$

$\displaystyle 0.028331382= \ln (1+r)$

Now take antilog,

$\displaystyle e^{0.028331382}= 1+r$

$\displaystyle 1.028736533= 1+r$

$\displaystyle 1.028736533-1= r$

$\displaystyle r=0.028736533$

$\displaystyle r = 2.87 \% $ - Sep 13th 2008, 01:47 PMSmoriginal
ahh I see. Thanks so much!!