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Thread: Logarithm help C2

  1. #1

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    Logarithm help C2

    can anybody help me with this question i am not to good a logarithms

    solve the equation

    log3 (2-3x) = log 9 (6x^2-19x+2)

    and

    if xy=64 and logxY+logyX=5/2 find x and y

  2. #2
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    Re: Logarithm help C2

    log3(2-3x) = log9(6x^2-19x+2) = log3(6x^2-19x+2)/log3(9) = log3(6x^2-19x+2)/2->
    2*log3(2-3x) = log3(6x^2-19x+2) -> log3((2-3x)^2) = log3(6x^2-19x+2) ->
    (2-3x)^2 = 6x^2-19x+2 -> 4 - 6x + 9x^2 = 6x^2 - 19x + 2 -> 3x^2+13x+2 = 0
    solving this equation you'll find 2 roots r1 and r2. 2-3x must be >= 0 and 6x^2-19x+2>=0. So a root r of the above is valid if 2-3r>=0 and 6r^2-19r+2>=0

    logxY+logyX = logyY/logyX+logyX=1/logyX+logyX=((logyX)^2+1)/logyX=5/2 ->
    2(logyX)^2 - 5logyX + 2 = 0 -> logyX = 2 or logyX = 1/2 ->
    logyX = logyY^2 or logyX^2 = logyY -> x = y^2 or y = x^2
    xy = 64 -> y^3 = 64 or x^3 = 64 -> y = 4 or x = 4 -> y=4 & x=16 or x=4 & y=16
    "bla, bla,... exists number M so for each n > M bla, bla..." Exists? Where is it? (Kronecker said...)

  3. #3
    vbuggy krtxmrtz's Avatar
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    Re: Logarithm help C2

    I thought I'd rewrite the first one in a tidier fashion.

    This type of problems can be solved using:

    logxy = log y / log x

    where log can be any base logarithm, particularly base 10.

    Then:

    log3(2 - 3x) = log9(6x2 - 19x +2)

    log(2 - 3x) / log 3 = log(6x2 - 19x +2) / log 9

    Now, because log 9 / log 3 = 2,

    2log(2 - 3x) = log(6x2 - 19x +2)

    or

    log(2 - 3x)2 = log(6x2 - 19x +2)

    and

    (2 - 3x)2 = 6x2 - 19x +2

    Rearranging and solving this 2-degree equation you get the correct solutions for x: -2 and -1/3
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  4. #4

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    Re: Logarithm help C2

    thanks alot that has really helped

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