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Math Help - Equation with square roots and logs

  1. #1
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    Equation with square roots and logs

    1) \sqrt{4x-7-12\sqrt{x-4}} + \sqrt{4x-16\sqrt{x-4}} = log_\frac{1}{2} \left ( x-\frac{25}{4} \right ) find x

    Waiting your answers

    repost your other questions one per thread
    Last edited by CaptainBlack; August 25th 2010 at 03:17 AM.
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  2. #2
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    Help me Please I need the answer
    This is from : Singapore Mathematics Olympiad 2001
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  3. #3
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    Quote Originally Posted by OnlyMath View Post
    1) \sqrt{4x-7-12\sqrt{x-4}} + \sqrt{4x-16\sqrt{x-4}} = log_\frac{1}{2} \left ( x-\frac{25}{4} \right ) find x

    Waiting your answers

    repost your other questions one per thread

    I think, there is no way to solve this monster !!!!!!!!!!!!!!!!!!!!
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  4. #4
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    I have the answer. But I want to know how to solve it!

    Can someone, a good mathematician, please show me the steps of how to solve it?

    Answer: x = 6.750...
    (Thanks to Wolfram|Alpha.)
    Wolfram alpha drew the graph of:

    \sqrt{4x-7-12\sqrt{x-4}} + \sqrt{4x-16\sqrt{x-4}} and log_\frac{1}{2} \left ( x-\frac{25}{4} \right ) then found where they intercepted to calculate what x was.

    Alternate form: \displaystyle{ln(x-25/4)+\sqrt{x-4 \sqrt{x-4}} ln(4)+\sqrt{4 x-12 \sqrt{x-4}-7} ln(2) = 0}

    I don't think you will get the answer anywhere else unless someone good at maths has a lot of time to solve it.
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  5. #5
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    Quote Originally Posted by OnlyMath View Post
    1) \sqrt{4x-7-12\sqrt{x-4}} + \sqrt{4x-16\sqrt{x-4}} = log_\frac{1}{2} \left ( x-\frac{25}{4} \right ) find x

    Waiting your answers
    \displaystyle\sqrt{4x-7-12\sqrt{x-4}} + \sqrt{4x-16\sqrt{x-4}} = log_{0.5} \left(x-\frac{25}{4}\right)

    \Rightarrow\displaystyle\ 0.5^{\left(\sqrt{4x-7-12\sqrt{x-4}} + \sqrt{4x-16\sqrt{x-4}}\right)}=x-6.25

    L=x-6.25

    L is a line, rising at 45 degrees, crossing the x-axis at x=6.25.

    y=0.5^{\left(\sqrt{4x-7-12\sqrt{x-4}} + \sqrt{4x-16\sqrt{x-4}}\right)}

    is positive, so L can only intersect it beyond x=6.25.

    y=0.5^{\left(\sqrt{4x-7-12\sqrt{x-4}} + \sqrt{4x-16\sqrt{x-4}}\right)}=0.5^{\left(\sqrt{4x-16\sqrt{x-4}+(4\sqrt{x-4}-7)}+\sqrt{4x-16\sqrt{4-x}}\right)}


    When x=6.25

    y=0.5^{\left(\sqrt{25-16\sqrt{2.25}+(4\sqrt{2.25}-7)}+\sqrt{25-16\sqrt{2.25}}\right)}

    and since

    15^2=225\Rightarrow\ (1.5)^2=2.25

    Or...

    \displaystyle\sqrt{2.25}=\sqrt{\frac{9}{4}}=\frac{  3}{2}

    x=6.25\Rightarrow\ y=0.5^{\left(\sqrt{18-16(1.5)+4(1.5)}}+\sqrt{25-16(1.5)}\right)}=0.5^{\left(\sqrt{24-24}+\sqrt{25-24}\right)}=0.5^{\left(\sqrt{0}+\sqrt{1}\right)}

    =0.5

    We can also evaluate the exponent easily when x=8.

    x=8\Rightarrow\ y=0.5^{\left(\sqrt{32-16(2)+4(2)-7}+\sqrt{32-16(2)}\right)}=0.5^{\left(\sqrt{1}+\sqrt{0}\right)  }=0.5


    Also, notice that if

    4\sqrt{x-4}-7=0\Rightarrow\ x-4=\frac{49}{16}\Rightarrow\ x=\frac{105}{16}=6.5625

    \Rightarrow\ 4x-16\sqrt{x-4}=\frac{49}{4}+16-16\,\left(\frac{7}{4}\right)=0.25

    x=6.5625\Rightarrow\ y=0.5^{\left(\sqrt{0.25}+\sqrt{0.25}\right)}=0.5^{  \left(0.5+0.5\right)}=0.5

    For what x will L be 0.5 ? and will y be 0.5 there too?
    Last edited by Archie Meade; August 28th 2010 at 02:07 PM.
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  6. #6
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    Hello, OnlyMath!

    I performed a lot of Olympic-level gymnastics,
    . . but it got me nowhere . . .


    \text{Solve for }x\!:

    . . \underbrace{\sqrt{4x-7-12\sqrt{x-4}}}_{A} + \underbrace{\sqrt{4x-16\sqrt{x-4}}}_{B} \;=\; \log_\frac{1}{2}\!\left ( x-\frac{25}{4} \right ) .[1]

    Under radical A we have: . 4x - 7 - 12\sqrt{x-4}

    Subtract and add 9: . 4x - 16 - 12\sqrt{x-4} + 9

    . . . . . . . . . . . . . =\; 4(x-4) - 12\sqrt{x-4} + 9

    . . . . . . . . . . . . . =\; 4(\sqrt{x-4})^2 - 12\sqrt{x-4} + 9

    . . . . . . . . . . . . . =\;(2\sqrt{x-4} - 3)^2

    Hence: . \text{Radical }A \;=\;\sqrt{4x - 7 -12\sqrt{x-4}} \;=\;2\sqrt{x-4} - 3



    Under radical B we have: . 4x - 16\sqrt{x-4}

    Subtract and add 16: . 4x-16 - 16\sqrt{x-4} + 16

    . . . . . . . . . . . . . =\;4\bigg[(x-4) - 4\sqrt{x-4} + 4\bigg]

    . . . . . . . . . . . . . =\;4\bigg[(\xqrt{x-4})^2 - 4\sqrt{x-4} + 4\bigg]

    . . . . . . . . . . . . . =\;4\bigg[\sqrt{x-4} - 2\bigg]^2

    Hence: . \text{Radical }B \;=\;\sqrt{4x-16\sqrt{x-4}} \;=\;2(\sqrt{x-4} - 2)



    Substitute into [1]: . (2\sqrt{x-4} - 3) + 2(\sqrt{x-4} - 2) \;=\;\log_{\frac{1}{2}}\!\!\left(x - \frac{25}{4}\right)


    And we have: . 4\sqrt{x-4} - 7 \;=\;\log_{\frac{1}{2}}\!\!\left(x-\frac{25}{4}\right)


    And there is no elementary solution to this transcendental equation.

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  7. #7
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    Quote Originally Posted by Soroban View Post
    Hello, OnlyMath!

    I performed a lot of Olympic-level gymnastics,
    . . but it got me nowhere . . .



    Under radical A we have: . 4x - 7 - 12\sqrt{x-4}

    Subtract and add 9: . 4x - 16 - 12\sqrt{x-4} + 9

    . . . . . . . . . . . . . =\; 4(x-4) - 12\sqrt{x-4} + 9

    . . . . . . . . . . . . . =\; 4(\sqrt{x-4})^2 - 12\sqrt{x-4} + 9

    . . . . . . . . . . . . . =\;(2\sqrt{x-4} - 3)^2

    Hence: . \text{Radical }A \;=\;\sqrt{4x - 7 -12\sqrt{x-4}} \;=\;2\sqrt{x-4} - 3



    Under radical B we have: . 4x - 16\sqrt{x-4}

    Subtract and add 16: . 4x-16 - 16\sqrt{x-4} + 16

    . . . . . . . . . . . . . =\;4\bigg[(x-4) - 4\sqrt{x-4} + 4\bigg]

    . . . . . . . . . . . . . =\;4\bigg[(\xqrt{x-4})^2 - 4\sqrt{x-4} + 4\bigg]

    . . . . . . . . . . . . . =\;4\bigg[\sqrt{x-4} - 2\bigg]^2

    Hence: . \text{Radical }B \;=\;\sqrt{4x-16\sqrt{x-4}} \;=\;2(\sqrt{x-4} - 2)

    Notice here that for x<8, you will have taken the negative square root of your square on the previous line...


    Substitute into [1]: . (2\sqrt{x-4} - 3) + 2(\sqrt{x-4} - 2) \;=\;\log_{\frac{1}{2}}\!\!\left(x - \frac{25}{4}\right)


    And we have: . 4\sqrt{x-4} - 7 \;=\;\log_{\frac{1}{2}}\!\!\left(x-\frac{25}{4}\right)


    And there is no elementary solution to this transcendental equation.

    Just one error there, Soroban.

    For 6.25<x<8...

    your Radical A is ok, but your Radical B should be

    2(2-\sqrt{x-4})

    The solution pops out nicely then..
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