Find the interest rate needed for an investment of $7,000 to grow to $10,500 in 6 yr if interest is compounded monthly. (Round your answer to the nearest hundredth of a percentage point.)

Printable View

- Nov 15th 2010, 11:25 AMdjdownfawlInterest Rate
Find the interest rate needed for an investment of $7,000 to grow to $10,500 in 6 yr if interest is compounded monthly. (Round your answer to the nearest hundredth of a percentage point.)

- Nov 15th 2010, 11:39 AMemakarov
You are supposed to show some effort. Do you know the formula for the total amount after n month? If yes, what is your difficulty?

- Nov 15th 2010, 11:46 AMdjdownfawl
a = p (1+ r/m) ^mt

?? - Nov 15th 2010, 01:15 PMemakarov
Yes, so $\displaystyle 10500=7000(1+r/12)^{12\cdot6}$, i.e., $\displaystyle (1+r/12)^{72}=1.5$. By raising both sides to 1/72, you can find r.

- Nov 17th 2010, 11:45 AMdjdownfawl
idk how to take the log and find r...

- Nov 17th 2010, 11:49 AMdjdownfawl
do u want me to take 72th root of both sides?

- Nov 17th 2010, 11:52 AMdjdownfawl
if i take the 72th root of both sides then r = 0 ...?

- Nov 17th 2010, 12:08 PMemakarovQuote:

if i take the 72th root of both sides then r = 0 ...?

- Nov 17th 2010, 12:09 PMdjdownfawl
ok i got it..

thanks the answer came out right..

r = 6.80 rounded - Nov 17th 2010, 12:11 PMdjdownfawl
But the other problem i am not able to figure out and i have 1 more try left its the last one..

can you please help - Nov 17th 2010, 12:11 PMdjdownfawl
Use logarithms to solve the equation for

*t*. http://www.webassign.net/cgi-bin/sym...%2F2%29%29%3DC - Nov 17th 2010, 12:32 PMemakarov
$\displaystyle \displaystyle\frac{A}{1+Be^{t/2}}=C$ iff

$\displaystyle A=C(1+Be^{t/2})$ and $\displaystyle 1+Be^{t/2}\ne0$.

$\displaystyle A=C(1+Be^{t/2})$ iff

$\displaystyle A=C+BCe^{t/2}$ iff

$\displaystyle A-C=BCe^{t/2}$ iff

$\displaystyle (A-C)/(BC)=e^{t/2}$ or $\displaystyle (A-C=0 and B=0)$.

$\displaystyle (A-C)/B=e^{t/2}$ iff

$\displaystyle \ln((A-C)/(BC))=t/2$. - Nov 17th 2010, 12:39 PMdjdownfawl
SO T = 2ln((A-C)/BC))

right? - Nov 17th 2010, 12:40 PMdjdownfawl
yessssssss its right!!!

thankkkk kyouuuuu soo muchhhh i appreciate your help very muchhhhh - Nov 17th 2010, 12:56 PMemakarov
You are welcome.