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 $a$ and $b$.

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

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

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

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
$f(x) = ax - \ln\left(\frac{x-b}{x+b}\right)$
$=ax -\ln(x-b)+\ln(x+b)$
then
$f'(x) = a -\frac{1}{x-b}+\frac{1}{x+b}$
Newton's method gives you an algorithm to find successive approximations to a root of the equation. If $a_n$ is the $n^{th}$approximation, then the next is given by:
$a_{n+1} = a_n - \frac{f(a_n)}{f'(a_n)}$
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.