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Math Help - Exponential Functions

  1. #1
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    Exponential Functions

    Investor A invests $1000 at an interest rate of 5% compounded continuously and
    investor B invests $500 at a rate of 8% compounded continuously. If both of
    them invest at the beginning of 2008, when will the value of their investments
    be equal.


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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Kblack View Post
    Investor A invests $1000 at an interest rate of 5% compounded continuously and
    investor B invests $500 at a rate of 8% compounded continuously. If both of
    them invest at the beginning of 2008, when will the value of their investments
    be equal.


    Kindly show details
    1000(1 + \frac{5}{100})^{n} = 500(1 + \frac{8}{100})^{n}

    2(1 + \frac{5}{100})^{n} = (1 + \frac{8}{100})^{n}

    2(\frac{21}{20})^n = (\frac{27}{25})^{n}

    2^{ \frac{1}{n} } = \frac{27}{25} \div \frac{21}{20}

    2^{ \frac{1}{n} } = \frac{36}{35}

    \frac{1}{n} = \log_{2} \frac{36}{35}

    I get n = 24,605

    So I'd say between 24 and 25 years.
    Last edited by janvdl; December 3rd 2007 at 07:58 AM. Reason: a spot of grammar error :D
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Kblack View Post
    Investor A invests $1000 at an interest rate of 5% compounded continuously and
    investor B invests $500 at a rate of 8% compounded continuously. If both of
    them invest at the beginning of 2008, when will the value of their investments
    be equal.

    Kindly show details
     <br />
1000(1.05)^n = 500(1.08)^n \implies 2 = \left( \frac{1.08}{1.05} \right)^n<br />

    \implies n = \dfrac{1}{\log_2 \left( \frac{1.08}{1.05} \right)} = \dfrac{\ln 2}{\ln \left( \frac{1.08}{1.05} \right)} \approx 24.6050977
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