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Math Help - Solving a logarithmic equation

  1. #1
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    Smile Solving a logarithmic equation

    Hi, could anyone help me with this, I'd be very greatful.Thanks..



    20logx^1/2 + 7logx^3 - 3logx^2 = 0
    4x 16x x/2
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  2. #2
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    Quote Originally Posted by alternative View Post
    Hi, could anyone help me with this, I'd be very greatful.Thanks..



    20logx^1/2 + 7logx^3 - 3logx^2 = 0
    4x 16x x/2
    hi

    This is confusing . You might want to try the latex (a software to format equations) .

    Try this ,

    logarithms of x to base 3 can be typed as \log_3(x) ,
    then wrap it with this \sum
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  3. #3
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    Quote Originally Posted by mathaddict View Post
    hi

    This is confusing . You might want to try the latex (a software to format equations) .

    Try this ,

    logarithms of x to base 3 can be typed as \log_3(x) ,
    then wrap it with this \sum



    Sorry about that, is it better now?

    Thanks alot
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  4. #4
    Junior Member mrmohamed's Avatar
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    Quote Originally Posted by alternative View Post


    Sorry about that, is it better now?

    Thanks alot
    HI all
    I think x will equal 4
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  5. #5
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    Using change of base (let new base be 2),the given problem can be written as
    10log(x)/log(4x) + 21log(x)/log(16x) = 6log(x)/log(x/2)
    Cancel log(x) from both side. You get
    10/log(4x) + 21/log(16x) = 6/log(x/2)
    10/[2 + log(x)] + 21/[4 + log(x) 6/[log(x) - 1]
    Let log x to the base 2 is a, then
    10/(2+a) + 21/(4+a) = 6/(a-1)
    Simplify this equation and solve for a. From that find x.
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  6. #6
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    Hello, alternative!

    20\log_{4x}\!\left(x^{\frac{1}{2}}\right) + 7\log_{16x}\!\left(x^3\right) - 3\log_{\frac{x}{2}}\!\left(x^2\right) \;=\;0

    Use the Base-Change formula and change everything to base-2 . . .

    . . . . . . \frac{20\log_2\left(x^{\frac{1}{2}}\right)}{\log_2  (4x)} + \frac{7\log_2(x^3)}{\log_2(16x)} - \frac{3\log_2(x^2)}{\log_2(\frac{x}{2})} \;=\;0

    . . \frac{20\cdot\frac{1}{2}\log_x(x)}{\log_2(x) + \log_2(4)} + \frac{7\cdot3\log_2(x)}{\log_2(x) + \log_2(16)} - \frac{3\cdot2\log(x)}{\log_2(x) - \log_2(2)} \;=\;0

    . . . . . . \frac{10\log_2(x)}{\log_2(x)+2} + \frac{21\log_2(x)}{\log_2(x) + 4} - \frac{6\log_2(x)}{\log_2(x)-1} \;=\;0


    Factor: . \log_2(x)\,\left[\frac{10}{\log_2(x)+2} + \frac{21}{\log_2(x)+4} - \frac{6}{\log_2(x) - 1}\right] \;=\;0



    Multiply through by the LCD. .
    I'll drop the "base-2" for now.

    . . \log(x)\,\bigg[10(\log x + 4)(\log x - 1) + 21(\log x + 2)(\log x - 1) - 6(\log x + 2)(\log x + 4)\bigg] \;=\;0

    . . . . . . . \log(x)\,\bigg[25\log^2x + 15\log x - 130\bigg] \;=\;0

    . . . . . . 5\log(x)\bigg[\log(x) - 2\bigg]\,\bigg[5\log(x) + 13\bigg] \;=\;0


    And we have three equations to solve:

    . . \log_2(x) \:=\:0\quad\Rightarrow\quad x \:=\:2^0 \quad\Rightarrow\quad \boxed{x \:=\:1}

    . . \log_2(x) \:=\:2 \quad\Rightarrow\quad x \:=\:2^2 \quad\Rightarrow\quad\boxed{x \:=\:4}

    . . 5\log_2(x) + 13 \:=\:0 \quad\Rightarrow\quad \log_2(x) \:=\:-\frac{13}{5} \quad\Rightarrow\quad \boxed{x \:=\:2^{-\frac{13}{5}}}

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