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Math Help - Logorythm equations

  1. #1
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    Question Logorythm equations

     lgx^{4} - \frac{30}{lgx} = 2

    I tried making lgx = y, but I can't get it to work.

    Next problem is:

     log_{2}x - log_{4}x + log_{16}x = 3

    I managed to get it down to  \frac{4}{4log_{x^{2}}} - \frac{2}{4log_{x^{2}}} + \frac{1}{4log_{x^{2}}} = 3

    but I don't know what next. The end result should be 16.

    And what about these two:

     \frac{z}{2 + lgx} + \frac{z}{4-lgx} = 1

    and

     log_{x}3 + log_{3}x = 2

    What I did, was  log_{x}3 + \frac{1}{log_{x}3} = 2

    Then,  log_{x}3 = y and that's where I'm stuck. What I get is:  y+\frac{1}{y}=2 I can't get a squared y out of it. And I don't know how to deal with the fraction.
    Please reply,
    Thanks!
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  2. #2
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    Re: Logorythm equations

    \log{x^4} - \frac{30}{\log{x}} = 2

    4\log{x} - \frac{30}{\log{x}} = 2

    4\log^2{x} - 30 = 2\log{x}

    2\log^2{x} - \log{x} - 15 = 0

    (2\log{x} + 5)(\log{x} - 3) = 0

    finish it ...


    \log_2{x} - \log_4{x} + \log_{16}{x} = 3

    use a change of base ...

    \log_2{x} - \frac{\log_2{x}}{\log_2{4}} + \frac{\log_2{x}}{\log_2{16}} = 3

    \log_2{x} - \frac{\log_2{x}}{2} + \frac{\log_2{x}}{4} = 3

    finish it ...
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  3. #3
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    Re: Logorythm equations

    Hello, dirtyharry!

    Here is the second one . . .


    \log_2x - \log_4x + \log_{16}x \:=\: 3

    Base-change Formula:

    n \frac{\log_2x}{\log_22} - \frac{\log_2x}{\log_24} + \frac{\log_2x}{\log_216} \;=\;3

    . . \frac{\log_2x}{1} - \frac{\log_2x}{2} + \frac{\log_2x}{4} \;=\;3

    . . . . . . . . . . . . . \tfrac{3}{4}\log_2x \;=\;3

    . . . . . . . . . . . . . . \log_2x \;=\;4

    . . . . . . . . . . . . . . . . . x \;=\;2^4 \;=\;16
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  4. #4
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    Re: Logorythm equations

    Quote Originally Posted by dirtyharry View Post
    ...

     log_{x}3 + log_{3}x = 2

    What I did, was  log_{x}3 + \frac{1}{log_{x}3} = 2

    Then,  log_{x}3 = y and that's where I'm stuck. What I get is:  y+\frac{1}{y}=2 ...
    Here is the last one:

    All your calculations are correct!

     y+\frac{1}{y}=2 ................. multiply through by y:

    y^2-2y+1=0~\implies~(y-1)^2=0

    The soultion of this equation should be obvious.

    Now solve for x:

     log_{x}3 = 1
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  5. #5
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    Re: Logorythm equations

    Quote Originally Posted by dirtyharry View Post
    ...

    And what about these two:

     \frac{z}{2 + lgx} + \frac{z}{4-lgx} = 1

    ...
    This one is the best!

    I assume that \lg(x) = \log_{10}(x)

     \frac{z(4-\lg(x))}{(2 + \lg(x))(4-\lg(x))} + \frac{z(2+\lg(x))}{(4-\lg(x))(2+\lg(x))} = 1

    Multiply through by the common denominator, expand the brackets and collect like terms at the LHS. You should come out with:

    (\lg(x))^2-2 \lg(x) + 6z -8 = 0 ............. Replace lg(x) by y:

    y^2-2y+6z-8=0 ............ Solve for y:

    y = 1 \pm \sqrt{9-6z}

    Therefore: z \le \frac32

    If

    z = \frac32~\implies~y = 1~\implies~x = 10

    z = \frac43~\implies~y = 2 \  \vee \  y=0 ~\implies~ x = 100 \  \vee \  x=1

    z = 0~\implies~y = 4 \  \vee \  y=-2 ~\implies~ x = 10000 \  \vee \  x=\frac1{100}

    etc.
    Last edited by earboth; December 12th 2012 at 11:21 AM.
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    Re: Logorythm equations

    Much appreciated
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