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Math Help - Economic Development - Logarithms problem

  1. #1
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    Economic Development - Logarithms problem

    Hi, I'm having a lot of difficulty with this problem because I'm terrible with logs.

    Consider two countries, Chairland and Couchia. In year 0, Chairland starts out with a per-capita
    income of $100, and this grows at a rate of 2% per year. Couchia starts out with a per-capita income
    of $1000, and this grows at a rate of 1% per year.
    In how many years will Chairland’s per-capita income catch up with Couchia’s per-capita income?
    (decimals are OK).

    So my setup was:
    x = years
    100 * 1.02^x = 1000 * 1.01^x
    = 1.02^x = 10 * 1.01^x

    I have no idea where to go from here. Could anyone show me what I should do next?
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  2. #2
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    Mar 2009
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    Quote Originally Posted by JoeZYe View Post
    So my setup was:
    x = years
    100 * 1.02^x = 1000 * 1.01^x
    = 1.02^x = 10 * 1.01^x
    =x \ln{1.02} = \ln {10}+x \ln {1.01}
    so;  x = \frac{\ln{10}}{\ln{1.02}-\ln {1.01}}=233.71


    Some important log rules;
    \ln{xy}=\ln{x}+\ln{y}
    \ln{\frac{x}{y}}=\ln{x}-\ln{y}
    \ln{x^{n}}=n\ln{x}

    So you could have said;
    \left( \frac{1.02}{1.01} \right)^{x}=10 and then taken logs and gotten the same answer...
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