# Interest Rate

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• Nov 15th 2010, 11:25 AM
djdownfawl
Interest 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 AM
emakarov
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 AM
djdownfawl
a = p (1+ r/m) ^mt
??
• Nov 15th 2010, 01:15 PM
emakarov
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 AM
djdownfawl
idk how to take the log and find r...
• Nov 17th 2010, 11:49 AM
djdownfawl
do u want me to take 72th root of both sides?
• Nov 17th 2010, 11:52 AM
djdownfawl
if i take the 72th root of both sides then r = 0 ...?
• Nov 17th 2010, 12:08 PM
emakarov
Quote:

if i take the 72th root of both sides then r = 0 ...?
No, if $\displaystyle (1+r/12)^{72}=1.5$, then $\displaystyle 1+r/12=(1.5)^{1/72}=\sqrt[72]{1.5}\approx1.00565$.
• Nov 17th 2010, 12:09 PM
djdownfawl
ok i got it..
thanks the answer came out right..
r = 6.80 rounded
• Nov 17th 2010, 12:11 PM
djdownfawl
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 PM
djdownfawl
Use logarithms to solve the equation for t. http://www.webassign.net/cgi-bin/sym...%2F2%29%29%3DC
• Nov 17th 2010, 12:32 PM
emakarov
$\displaystyle \displaystyle\frac{A}{1+Be^{t/2}}=C$ 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 PM
djdownfawl
SO T = 2ln((A-C)/BC))

right?
• Nov 17th 2010, 12:40 PM
djdownfawl
yessssssss its right!!!

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