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Math Help - exponentials and logarithms

  1. #1
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    exponentials and logarithms

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    3^x = 4^y = 12^z,

    that z = xy/(x+y)

    Tried looking at 3^x = 4^y and taking log base 12 to eliminate one of the variables, ending up with x=ylog base 3 (4).

    then I looked at 4^y = 12^z, took log base 12 as well, and ended up with z= y*log base 12 (4).

    Plugged x and y into xy/(x+y), but couldnt get it to equal z

    had a fiddle about with it, but I can't seem to prove it. maybe, im thinking too hard, anybody else got ideas?
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  2. #2
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    3^x = 4^y = 12^z = k

     3 = k^1/x, 4 = k^1/y and 12 = k^1/z

    12 = 3*4

    Substitute the values and simplify.
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  3. #3
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    3^x = 4^y = 12^z = k

     3 = k^{1/x}, 4 = k^{1/y} and 12 = k^{1/z}

    12 = 3*4

    Substitute the values and simplify.
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  4. #4
    MHF Contributor Amer's Avatar
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    Jordan
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    from

    12^z = 4^y find z value

     z = \log_{12} 4^y

    z = y \log_{12} 4 same thing from 12^z = 3^x

    you will have

    z = y \log_{12} 4
    z = x \log_{12} 3

    multiply first one with x and the second with y

    zx = yx \log_{12} 4
    zy = yx \log_{12} 3

    add them

     zx + zy = yx( \log_{12} 4 + \log_{12} 3 )

    z (x+y) = yx (\log_{12} 3(4) )

    z = \frac{xy}{x+y}
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  5. #5
    MHF Contributor Amer's Avatar
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    or u can find y,x values with respect to z then sub them in xy/(x+y) u should get z
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  6. #6
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    thanks guys.

    got it!
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