What 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

Quote:

Originally Posted by Smoriginal

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

yes you have to use ln, see here,

$<br /> \frac{2773}{1000}=(1+r)^{36}$

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

Now take natural log on both sides:

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

$<br /> 1.019929767= 36 \ln (1+r)$

$<br /> \frac{1.019929767}{36}= \ln (1+r)$

$0.028331382= \ln (1+r)$

Now take antilog,

$e^{0.028331382}= 1+r$

$1.028736533= 1+r$

$1.028736533-1= r$

$r=0.028736533$

$r = 2.87 \%$

ahh I see. Thanks so much!!