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Thread: Equation

  1. #1
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    Equation

    a*x=ln{(x-b)/(x+b)}
    a and b are constants.
    Any help or idea to solve the equation.
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  2. #2
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    Hello Emilly
    Quote Originally Posted by Emilly View Post
    a*x=ln{(x-b)/(x+b)}
    a and b are constants.
    Any help or idea to solve the equation.
    You'll have to use a numerical method or something like a spreadsheet to solve this equation, and it will only have a solution for certain values of $\displaystyle a$ and $\displaystyle b$.

    I attach a graph I produced in Excel, to show the result when $\displaystyle a = -0.1$ and $\displaystyle b = 10$. For these values, the equation has the approximate solution $\displaystyle x = 15.43$, which I found this value using Excel's 'Goal Seek'. There is also another branch of the log graph for values of $\displaystyle x$ less than $\displaystyle -10$, which I haven't included in the diagram.

    You'll see that there are no points on the log graph where $\displaystyle -b\le x\le b$, and as $\displaystyle x\to \infty$, the graph $\displaystyle \to 0$ from below. So there won't be any solution if $\displaystyle a \ge 0$ for positive values of $\displaystyle b$.

    I'll leave you to work out what will happen to the graphs for different values of $\displaystyle a$ and $\displaystyle b$.

    Grandad
    Attached Thumbnails Attached Thumbnails Equation-untitled.jpg  
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  3. #3
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    mm I need some algorithm.
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  4. #4
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    Hello Emilly
    Quote Originally Posted by Emilly View Post
    mm I need some algorithm.
    Do you know Newton's method (also known as Newton-Raphson) for the approximate solution of equations?

    If
    $\displaystyle f(x) = ax - \ln\left(\frac{x-b}{x+b}\right)$
    $\displaystyle =ax -\ln(x-b)+\ln(x+b)$
    then
    $\displaystyle f'(x) = a -\frac{1}{x-b}+\frac{1}{x+b}$
    Newton's method gives you an algorithm to find successive approximations to a root of the equation. If $\displaystyle a_n$ is the $\displaystyle n^{th} $approximation, then the next is given by:
    $\displaystyle a_{n+1} = a_n - \frac{f(a_n)}{f'(a_n)}$
    But if all you need is a numerical answer to a particular equation, then it's much quicker to use a 'Goal Seek' method in a spreadsheet.

    Grandad
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