# Compound interest problem

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• Dec 18th 2008, 08:13 AM
mwok
Compound interest problem
Question:
In 1974, Johnny Miller won 8 tournaments on the PGA tour and accumulated $304,680 in official season earnings. In 1999, Tiger Woods accumulated$6081912 with a similar record. If Miller had invested his earnings in an account earning compound interest, find the annual interest rate needed for his winnings to be equivalent in value to Tiger Wood's in 1999.

My solution is attached however it's not complete. I don't know how to get ride of the (1 + r)^25 exponent.
• Dec 18th 2008, 08:40 AM
JD-Styles
Take the 25th root of both sides:

$19.96^{1/25}=1+r$
• Dec 18th 2008, 08:47 AM
mwok
Okay, so after solving the problem r = 0.12

I'm not sure that is correct. Can someone confirm if it is correct?
Here is the solution:
• Dec 18th 2008, 08:48 AM
Soroban
Hello, mwok!

You were doing great . . .

Quote:

$304,\!680(1+r)^{25} \:=\:6,\!081,\!912$

$(1+r)^{25} \:\approx\:19.96$

Take the $25^{th}$ root of both sides:

. . $\left[(1+r)^{25}\right]^{\frac{1}{25}} \;=\;(19.96)^{\frac{1}{25}}$

. . . . . . $1 + r \;=\;1.127217824$

. . . . . . . . $r \;=\;0.127217824 \;\approx\;12.7\%$

Ah, too slow . . . again!
.
• Dec 18th 2008, 08:54 AM
fonso_gfx
Compound interest problem
Hi, Mwork

You were very close to the answer, you just have to raise both sides to the 1/25 power.

$
\frac{6081912}{304680} = (1+r)^{25}
$

$(\frac{6081912}{304680})^{\frac{1}{25}} = (1+r)$

$(\frac{6081912}{304680})^{\frac{1}{25}}-1 = r$

Next time use this formula for the rate

$
r = (\frac{A}{P})^{\frac{1}{n}}-1
$
• Dec 18th 2008, 08:57 AM
mwok
Thanks, The question is asking for the rate. Should I leave it as 0.12 or 12.7? Is it correct if I left it as 0.12?
• Dec 18th 2008, 10:17 AM
tester85
It is best to leave it as 12.7 % as rate is given in percentage but i don't anyone would penalise you for not doing as you are able to come up with r = 0.127 (Nod)
• Dec 19th 2008, 07:47 PM
Tim28
(1+i)^25=19.961630
take log of both sides,
25log(1+i)=log 19.961630=1.300196
log(1+i)=1.300196/25=.052
1+i=10^.052=1.1272, so i=12.72