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Math Help - solve for x

  1. #1
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    solve for x

    Solve x+20ln|x-20|=234

    The given answer is x is approximately 139, if you need to check.
    Last edited by Punch; December 19th 2011 at 05:50 AM.
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  2. #2
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    Re: solve for x

    Do you know about the Lambert W Function?

    \begin{align*} x+20 \cdot \ln(x-20) &=234 \\ 20\cdot \ln(x-20)&=234-x \\ \ln(x-20)&=\frac{234-x}{20} \\ x-20 &= e^{\frac{234-x}{20}} \\ 20\left( \frac{x}{20}-1\right) &= {e^{117 \over 10} \over e^{x \over 20}} \\  \frac{x}{20}-1 &= {e^{117 \over 10} \over 20 \cdot e^{x \over 20}} \\  \frac{x}{20}-1 &= {e^{107 \over 10}\cdot e \over 20 \cdot e^{x \over 20}}  \\  \frac{x}{20}-1 &= {e^{107 \over 10} \over 20 \cdot e^{{x \over 20}-1}} \\   e^{{x \over 20}-1}\left( \frac{x}{20}-1\right) &= {e^{107 \over 10}\over  20} \\ {x \over 20}-1 &= W{\left( {e^{107 \over 10}\over  20}\right)} \\ x &=  20\left(  W{\left( {e^{107 \over 10}\over  20}\right)}+1\right) \\ x &\approx 138.501\end{align*}
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  3. #3
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    Re: solve for x

    Quote Originally Posted by sbhatnagar View Post
    Do you know about the Lambert W Function?

    \begin{align*} x+20 \cdot \ln(x-20) &=234 \\ 20\cdot \ln(x-20)&=234-x \\ \ln(x-20)&=\frac{234-x}{20} \\ x-20 &= e^{\frac{234-x}{20}} \\ 20\left( \frac{x}{20}-1\right) &= {e^{117 \over 10} \over e^{x \over 20}} \\  \frac{x}{20}-1 &= {e^{117 \over 10} \over 20 \cdot e^{x \over 20}} \\  \frac{x}{20}-1 &= {e^{107 \over 10}\cdot e \over 20 \cdot e^{x \over 20}}  \\  \frac{x}{20}-1 &= {e^{107 \over 10} \over 20 \cdot e^{{x \over 20}-1}} \\   e^{{x \over 20}-1}\left( \frac{x}{20}-1\right) &= {e^{107 \over 10}\over  20} \\ {x \over 20}-1 &= W{\left( {e^{107 \over 10}\over  20}\right)} \\ x &=  20\left(  W{\left( {e^{107 \over 10}\over  20}\right)}+1\right) \\ x &\approx 138.501\end{align*}
    Sorry but I dont know the Lambert W Function? Is there another way to solve this?
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  4. #4
    Member sbhatnagar's Avatar
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    Re: solve for x

    Quote Originally Posted by Punch View Post
    Sorry but I dont know the Lambert W Function? Is there another way to solve this?
    You can use graphical method.

    Plot y=x+20ln(x-20) and y=234

    Note the point where they intersect. The abscissa of the point is the value of x.

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  5. #5
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    Re: solve for x

    Quote Originally Posted by sbhatnagar View Post
    You can use graphical method.

    Plot y=x+20ln(x-20) and y=234

    Note the point where they intersect. The abscissa of the point is the value of x.

    Yeap, I tried that but the graph of y=x+20ln(x-20) seems totally out of range on my graphical calculator
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  6. #6
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    Re: solve for x

    Quote Originally Posted by Punch View Post
    Yeap, I tried that but the graph of y=x+20ln(x-20) seems totally out of range on my graphical calculator
    Since you didn't tell us which model you use I've attached a telenovela taken from a TI84. Graph the function:

    y=x+20\cdot\ln|x-20|-234

    and determine the x-intercept.

    Maybe this helps.
    Attached Thumbnails Attached Thumbnails solve for x-graphlsg.png  
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