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    Hey if anyone could help with the following:

    1) Solve:
    2log3x - log3 (x-2) = 2 x>2


    2) Every £1 of money invested in a saving scheme continuously gains interest at a rate of 4% per year. Hence after x years, the total value of an initial £1 investment is £y, where

    y=1.04^x

    Use logarithms to find the number of years it takes to double the total value of any initial investment.

    Thanx =]
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    1) Can be rewritten as
    log3x^2-log3(x-2)=2 which can be arranged to
    log3x^2/(x-2)=2 [logax-logbx=loga/b]
    x^2/(x-2) = 3^2 = 9
    x^2 = 9x-18
    x^2-9x+18= 0
    (x-6)(x-3)=0
    x = 6 or 3
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    (Original post by samanthaaaa)
    1) Can be rewritten as
    log3x^2-log3(x-2)=2 which can be arranged to
    log3x^2/(x-2)=2 [logax-logbx=loga/b]
    x^2/(x-2) = 3^2 = 9
    x^2 = 9x-18
    x^2-9x+18= 0
    (x-6)(x-3)=0
    x = 6 or 3
    Thankyou!
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    for 2)

    take logs of both sides
    log(y)=log(1.04^x)
    drop the x down using power rules
    log(y)=xlog(1.04)

    so x=log(y)/log(1.04)
    you're looking for when y=2 (double initial value)
    so x=log(2)/log(1.04)

    put this in a calculator, doesn't matter what base, and you'll get x=17.7 to 3sf
 
 
 
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