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Math Help - log 3

  1. #1
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    log 3

    1. If log_{9}5=x and log_{27}6=y, find an expression in terms of x and y for log_{3}1.2.
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  2. #2
    MHF Contributor red_dog's Avatar
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    Transform the logarithms in base 3:

    x=\frac{\log_3 5}{2}, \ y=\frac{\log_3 2+1}{3}

    \log_3(1.2)=\log_3\frac{6}{5}=\log_3 2+1-\log_3 5=3y-2x
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  3. #3
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    I would have done this a slightly different way, using exponentials rather than logarithms. log_9(5)= x so 9^x= 5. log_{27}(6)= y so 27^y= 6.
    9= 3^2 so 9^x= (3^2)^x= 3^{2x}= 5 and 27= 3^3 so 27^y= (3^3)^y= 3^{3y}= 6.

    Since 1.2= 6/5, we have 1.2= \frac{6}{5} = \frac{3^{3y}}{3^{2x}} = 3^{3y-2x} and so log_3(1.2)= 3y- 2x.
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