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Math Help - Investment interest problem

  1. #1
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    Investment interest problem

    "An investment of $1000 earns $177.23 in interest in two years. If the interest is compounded annually, find the annual interest rate. Round to the nearest tenth of a percent.)

    Using the formula P = A(1 + i)^2, I solved it thus:

    1177.23 = 1000(1 + i)^2

    1.17723 = (1 + i)^2

    log 1.17723 = 2 log (1 + i)

    log 1.17723 = 2 log 1 + 2 log i

    \frac {log 1.17723}{2} = log i

    .0354306605 = log i

    10^{.0354306605} = i

    i \approx 1.085

    As it turns out, my answer is correct if you remove the integer portion and multiply by 100, i.e. 8.5%. However, the answer I get is actually 108.5%. Where did I go wrong?

    The book gives a hint... "The solution will require taking the common logarithm of each side of the equation and taking the common antilogaritm of each side of the equation."... but I'm not quite understanding how to do that.

    Thanks in advance for your help!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by earachefl View Post
    "An investment of $1000 earns $177.23 in interest in two years. If the interest is compounded annually, find the annual interest rate. Round to the nearest tenth of a percent.)

    Using the formula P = A(1 + i)^2, I solved it thus:

    1177.23 = 1000(1 + i)^2

    1.17723 = (1 + i)^2

    log 1.17723 = 2 log (1 + i)

    log 1.17723 = 2 log 1 + 2 log i

    \frac {log 1.17723}{2} = log i

    .0354306605 = log i

    10^{.0354306605} = i

    i \approx 1.085

    As it turns out, my answer is correct if you remove the integer portion and multiply by 100, i.e. 8.5%. However, the answer I get is actually 108.5%. Where did I go wrong?

    The book gives a hint... "The solution will require taking the common logarithm of each side of the equation and taking the common antilogaritm of each side of the equation."... but I'm not quite understanding how to do that.

    Thanks in advance for your help!
    the formula you used is not to find the interest only, it is to find the principal plus the interest.

    the formula you should use to find only the interest is: I = (1 + r)^t ...where I is the interest gained

    you followed the hints given correctly
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  3. #3
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    OK, if I use your formula:

    I = (1 + r)^2, I get:

    177.23 = (1 + r)^2

    log 177.23 = 2 log (1 + r)

    log 177.23 = 2 log 1 + 2 log r

    \frac {log 177.23}{2} = log r

    1.1242686187 = log r

    10^{1.1242686187} = r

    r \approx 13.3

    Unfortunately, that's not the correct answer, as you can see if you plug 13.3% back into the original formula:
    P = 1000(1 + .133)^2
    P = 1283.69

    Did I misunderstand you?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by earachefl View Post
    OK, if I use your formula:

    I = (1 + r)^2, I get:

    177.23 = (1 + r)^2

    log 177.23 = 2 log (1 + r)

    log 177.23 = 2 log 1 + 2 log r

    \frac {log 177.23}{2} = log r

    1.1242686187 = log r

    10^{1.1242686187} = r

    r \approx 13.3

    Unfortunately, that's not the correct answer, as you can see if you plug 13.3% back into the original formula:
    P = 1000(1 + .133)^2
    P = 1283.69

    Did I misunderstand you?
    no, it was my bad. ....forget what i said!

    your formula is correct, you made a mistake with the logs. i'll see if i find it

    here

    Quote Originally Posted by earachefl View Post
    "An investment of $1000 earns $177.23 in interest in two years. If the interest is compounded annually, find the annual interest rate. Round to the nearest tenth of a percent.)

    Using the formula P = A(1 + i)^2, I solved it thus:

    1177.23 = 1000(1 + i)^2

    1.17723 = (1 + i)^2

    log 1.17723 = 2 log (1 + i)

    log 1.17723 = \color{red} 2 log 1 + 2 log i
    here is your mistake. you cannot distribute the log among a sum like that.

    the next line should be: \frac {\log 1.17723}2 = \log (1 + i)

    now continue
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  5. #5
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    Wouldn't

    <br />
\begin{aligned}<br />
1000x^{2}&=1177.23\\\\<br />
x^{2}&=\frac{1177.23}{1000}\\\\<br />
x&=\sqrt{\frac{1177.23}{1000}}\\\\<br />
x&\approx{1.08500}<br />
\end{aligned}<br />

    <br />
1.08500-1=0.08500<br />

    be easier? Of course,

    <br />
0.085 = 8.5\%<br />
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by p.numminen View Post
    Wouldn't

    <br />
\begin{aligned}<br />
1000x^{2}&=1177.23\\\\<br />
x^{2}&=\frac{1177.23}{1000}\\\\<br />
x&=\sqrt{\frac{1177.23}{1000}}\\\\<br />
x&\approx{1.08500}<br />
\end{aligned}<br />

    <br />
1.08500-1=0.08500<br />

    be easier? Of course,

    <br />
0.085 = 8.5\%<br />
    indeed. the use of logs was not necessary here (we use logs when we have unknowns as powers and regular manipulation of the exponents does not work). however, we could have left x as (1 + i) just to show where the -1 in the second to last equation comes from
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