# Thread: Compound interest problem

1. ## 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.

2. Take the 25th root of both sides:

$19.96^{1/25}=1+r$

3. 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:

4. Hello, mwok!

You were doing great . . .

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

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

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!
.

5. ## 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
$

6. 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?

7. 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

8. (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