1. Equation

a*x=ln{(x-b)/(x+b)}
a and b are constants.
Any help or idea to solve the equation.

2. Hello Emilly
Originally Posted by Emilly
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 $\displaystyle b$.

3. mm I need some algorithm.

4. Hello Emilly
Originally Posted by Emilly
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.