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Math Help - [SOLVED] Consumer Mathematics

  1. #1
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    [SOLVED] Consumer Mathematics

    I placed $1200 in an bank investment which paid 8% p.a compound interest. At the same time, I placed $1800 in a bank which paid 4% p.a compound interest. During which year would the $1200 investment be worth more than the $1800 investment. I can do this question by listing the figures at the end of every year, but I wanted to know if there is way to solve this using algebra?
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  2. #2
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    Compound Interest

    Hello requal
    Quote Originally Posted by requal View Post
    I placed $1200 in an bank investment which paid 8% p.a compound interest. At the same time, I placed $1800 in a bank which paid 4% p.a compound interest. During which year would the $1200 investment be worth more than the $1800 investment. I can do this question by listing the figures at the end of every year, but I wanted to know if there is way to solve this using algebra?
    The compound interest formula gives the amount A after n years as:

    A = P(1+r)^n

    So into this formula you have to plug in each of the values of P and r, equate the two and solve for n to find the point at which the two amounts become equal.

    1200(1.08)^n = 1800(1.04)^n

    \Rightarrow 1.08^n = 1.5(1.04)^n

    \Rightarrow \left(\frac{1.08}{1.04}\right)^n = 1.5

    \Rightarrow 1.03846^n = 1.5

    Now - and here's the key step - take logs of both sides (it doesn't matter which base of logs you use):

    n\log(1.03846) = \log(1.5)

    \Rightarrow n = \frac{\log(1.5)}{\log(1.03846)}

    = 10.74

    So by the end of year 11, the first investment will be worth more than the second.

    Grandad
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