Interest Rate

**djdownfawl** · November 15th 2010, 11:25 AM

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.)

**emakarov** · November 15th 2010, 11:39 AM

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?

**djdownfawl** · November 15th 2010, 11:46 AM

a = p (1+ r/m) ^mt
??

**emakarov** · November 15th 2010, 01:15 PM

Yes, so $10500=7000(1+r/12)^{12\cdot6}$ , i.e., $(1+r/12)^{72}=1.5$ . By raising both sides to 1/72, you can find r.

**djdownfawl** · November 17th 2010, 11:45 AM

idk how to take the log and find r...

**djdownfawl** · November 17th 2010, 11:49 AM

do u want me to take 72th root of both sides?

**djdownfawl** · November 17th 2010, 11:52 AM

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

**emakarov** · November 17th 2010, 12:08 PM

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

No, if $(1+r/12)^{72}=1.5$ , then $1+r/12=(1.5)^{1/72}=\sqrt[72]{1.5}\approx1.00565$ .

**djdownfawl** · November 17th 2010, 12:09 PM

ok i got it..
thanks the answer came out right..
r = 6.80 rounded

**djdownfawl** · November 17th 2010, 12:11 PM

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

**djdownfawl** · November 17th 2010, 12:11 PM

Use logarithms to solve the equation for t.

**emakarov** · November 17th 2010, 12:32 PM

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

$A=C(1+Be^{t/2})$ iff
$A=C+BCe^{t/2}$ iff
$A-C=BCe^{t/2}$ iff
$(A-C)/(BC)=e^{t/2}$ or $(A-C=0 and B=0)$ .

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

**djdownfawl** · November 17th 2010, 12:39 PM

SO T = 2ln((A-C)/BC))

right?

**djdownfawl** · November 17th 2010, 12:40 PM

yessssssss its right!!!

thankkkk kyouuuuu soo muchhhh i appreciate your help very muchhhhh

**emakarov** · November 17th 2010, 12:56 PM

You are welcome.

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