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Thread: Square root in a logarithmic equation

  1. #1
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    Square root in a logarithmic equation

    Square root in a logarithmic equation-untitled.jpg

    Hello,

    I am having difficulty isolating x with the square root involved in this equation. I combine the left side with the property of logs that subtraction is equivalent to division, then make both sides an exponent with base 10 to eliminate the log. I then cross multiplied 7x/(9+sqrt(x)) and 100/1. I am then stuck with 900+100(sqrt(x))=7x.

    Any help would be greatly appreciated.

    Thank you!
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  2. #2
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    Re: Square root in a logarithmic equation

    Quote Originally Posted by mathguy5000 View Post
    Click image for larger version. 

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    base 10 to eliminate the log. I then cross multiplied 7x/(9+sqrt(x)) and 100/1. I am then stuck with 900+100(sqrt(x))=7x.
    $\begin{align*}7x&=9\cdot10^2+10^2\sqrt{x} \\(7x- 9\cdot10^2)^2&=10^4x \end{align*}$
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  3. #3
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    Re: Square root in a logarithmic equation

    Quote Originally Posted by mathguy5000 View Post
    I am then stuck with 900+100(sqrt(x))=7x.
    sqrt(x) = (7x - 900) / 100

    Solve for x by squaring both sides.
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    Re: Square root in a logarithmic equation

    Ahh. Thank you very much!
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    Re: Square root in a logarithmic equation

    Does this then become a quadratic equation? I have the final answer in a key, as this is a practice question for my class, but I have tried time and time again and cannot get it to work through to match the solution for the numerical value of x.
    Last edited by mathguy5000; Feb 21st 2017 at 03:45 PM.
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    Re: Square root in a logarithmic equation

    Quote Originally Posted by mathguy5000 View Post
    Does this then become a quadratic equation? I have the final answer in a key, as this is a practice question for my class, but I have tried time and time again and cannot get it to work through to match the solution for the numerical value of x.
    HERE is a solution.
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    Re: Square root in a logarithmic equation

    $900+100\sqrt{x}=7x$

    Squaring both sides

    $(900+100\sqrt{x})^2=(7x)^2$

    $810000+10000x^2+180000x=49x^2$

    $9951x^2+180000x+810000=0$

    It's a quadratic equation so, solve it using quadratic formula.

    $x=\dfrac{-180000\pm\sqrt{(180000)^2-4\times9951\times810000}}{2\times9951}$

    $x=-\dfrac{300}{31}$

    $x=-\dfrac{900}{107}$
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  8. #8
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    Re: Square root in a logarithmic equation

    Quote Originally Posted by deesuwalka View Post
    $900+100\sqrt{x}=7x$

    Squaring both sides

    $(900+100\sqrt{x})^2=(7x)^2$
    OK so far so ...

    Quote Originally Posted by deesuwalka View Post
    $810000+10000x^2+180000x=49x^2$
    Nope - this should be:

    $81000 + 10000x +180000\sqrt x = 49 x^2$

    which is not much help. And of course your final answers are both negative, which can't be correct.

    Better to do as Plato and DenisB have suggested.

    To Mathguy500: the quadratic that you need to solve comes from squaring 7x-900 = 100\sqrt x, to get:

    49x^2 - 12600 x +810000 = 10000x

    which leads to:

     49x^2 - 22600 x +81000 = 0

    Now you can apply the quadratic equation. You'll get two results - you'll have to try both back in the original equation to see which one is correct.
    Last edited by ChipB; Feb 22nd 2017 at 03:54 AM.
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