Stuck at the end of an interest problem involving ln

larasanut · #1 $Old$ December 3rd, 2008, 02:01 PM

in short..the problem is 7500=5000(1+r/12)^36

i'm down to this part e^.01126 = (1+r/12)

where do i go from here?

skeeter · #2 $Old$ December 3rd, 2008, 03:11 PM

Quote:

7500=5000(1+r/12)^36

I guess you're solving for "r" ?

$1.5 = \left(1 + \frac{r}{12}\right)^{36}$

$1.5^{\frac{1}{36}} = 1 + \frac{r}{12}$

$r = 12\left(1.5^{\frac{1}{36}} - 1\right) \approx 13.6$ %

larasanut · #3 $Old$ December 3rd, 2008, 03:51 PM

thats the right answer but my book does it differently...but at the very end it loses me...

the book does (i'm going to use the book example)

18000=10000(1+r/12)^12(5)
= 1.8 = (1+r/12)60
then use logarithm

ln 1.8=ln(1+r/12)60
=ln 1.8 = 60ln(1+r/12)
=ln1.8/60 = (1+r/12)
=.009796 = (1+r/12)

=(1+r/12) = e^.009796
= 1.009844 <--------

here's where i get stuck...where did they get 1.009844

and then it continues to r/12=1.009844-1 =.1181 also 11.81%

u're answer is right, i just want to understand where they got the 1.00944 from e^.009796?

skeeter · #4 $Old$ December 3rd, 2008, 04:01 PM

larasanut · #5 $Old$ December 3rd, 2008, 04:07 PM

ahh, i got it..i wasn't even thinking about using my calculator thanks