How to solve logarithmic functions I'd like to know is there any way can solve the equation such as ln(1+x) = ax+b, where ln() is natural log. I can plot the figure out and find the intersection, but now 'a' and 'b' are arbitary numbers and is there any approximation approach to estimate the solution?
 A: We can do this using the Lambert W function, the inverse of 
$$f(x) = xe^x$$
We have
$$\ln(1+x)=ax+b$$
$$1+x = e^{ax+b}$$
Substituting $y=1+x$:
$$y=e^{ay-a+b}$$
$$ye^{-ay} = e^{b-a}$$
$$-aye^{-ay} = -a e^{b-a}$$
$$-ay = W\left(-ae^{b-a}\right)$$
$$x = -\frac{W\left(-ae^{b-a}\right)}{a}-1$$
Note that we can also use $W_{-1}$, the other branch of the $W$ function, giving two possible answers. 
A: Usually these are solved by Newton type methods, i.e. 
$$f(x) = ln(1+x) - ax - b$$
$$f'(x) = \frac{1}{1+x} - a$$
Start with an initial guess $x_0$, then iterate as follows:
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{ln(1+x_n) - ax_n - b}{\frac{1}{1+x_n} - a}$$
A: El Bazzi gives one approximation technique.  There is a similar one called "Secant Line Method" which avoids calculus.  Using the same function $f(x)$ you pick two values of $x_1$, $x_2$ which are (roughly) close to the solution.  Plug them into the function to get two points.  Find the equation of the line through those two points.  Let $x_3$ be the $x$-intercept of that line.  $x_3$ should be a closer approximation to the solution.  Then repeat using the two points $x_2$ and $x_3$ to get $x_4$.  The sequence $x_1, x_2, x_3, \ldots$ should converge to the solution.  (But the method is not always stable.  Unlucky choices of $x_1$ and $x_2$ may give a divergent sequence.)
