# Solution of equation of form ln(a+bx) = cx

• Apr 17th 2012, 08:52 AM
fraenzine
Solution of equation of form ln(a+bx) = cx
Hello,

I am having an exponential equation that depends on 3 different variables, which are included in the constants a, b and c....
solving this equation, I get something of the form of ln(a + bx) - cx = 0

I don't manage to simplify the equation any further (Worried)
I was wondering whether this is actually possible without any numerical method, I would like to get an equation for x as a function of a, b and c (or actually some other constants that form a, b and c).

Or is there maybe any approximations method , such as a Taylor development or something, which gives me a solution with a certain accuracy?

Would be nice if someone could help me!

Cheers (Wink)
• Apr 17th 2012, 10:18 AM
Deveno
Re: Solution of equation of form ln(a+bx) = cx
i think you need to use the Lambert W-function here: Lambert W-Function -- from Wolfram MathWorld
• Apr 18th 2012, 03:56 AM
fraenzine
Re: Solution of equation of form ln(a+bx) = cx

I had already heard about the Lambert W-function in order to solve problems like this, but I have to admit that I don't really understand how that works... (I'm not tha good in maths, I'm just a chemist ;) )
I understood what the Lambert function is but I don't manage to apply it to solve my problem :S

Thanks a lot ;)
• Apr 18th 2012, 04:15 AM
princeps
Re: Solution of equation of form ln(a+bx) = cx
• Apr 18th 2012, 04:51 AM
a tutor
Re: Solution of equation of form ln(a+bx) = cx
$\displaystyle \displaystyle a+bx=e^{cx}$

$\displaystyle \displaystyle (a+bx)e^{-cx}=1$

$\displaystyle \displaystyle -(\frac{ac}{b}+cx)e^{-cx}=-\frac{c}{b}$

$\displaystyle \displaystyle -(\frac{ac}{b}+cx)e^{-(ac/b+cx)}=-\frac{c}{b}e^{-ac/b}$

This is now ready to apply the Lambert function...

Pretty sure I've got a slip in there somewhere so look out.
• Apr 18th 2012, 07:22 AM
fraenzine
Re: Solution of equation of form ln(a+bx) = cx
Thanks a lot :)
But I still don't know how to APPLY the Lambert-W function... and so how to get a solution for x....
• Apr 18th 2012, 11:54 AM
MathoMan
Re: Solution of equation of form ln(a+bx) = cx
Dude forget that Lambert W(hatever) stuff.

Here goes:

First plot the functions g(x)=ln(a+bx) and h(x)=cx, both in the same coordinate plane. Abscissas of the intersection points are solutions to the equation g(x)=h(x) or ln(a+bx)=cx. Now from that picture you can estimate the intervals that contain the solutions. Say that interval is I=[u,v].

Now you just have state the precision you need (epsilon=0.00005 would mean 4 significant decimal digits) and use an iterative method that would find the solution in the interval I=[u,v]. Good choice of method would be Newton tangent method, or even plain bisection method provided you don't need great precision.
• Apr 18th 2012, 01:11 PM
a tutor
Re: Solution of equation of form ln(a+bx) = cx
$\displaystyle \displaystyle W\left(-\left(\frac{ac}{b}+cx\right)e^{-(ac/b+cx)}\right)=W\left(-\frac{c}{b}e^{-ac/b}\right)$

$\displaystyle \displaystyle -\left(\frac{ac}{b}+cx\right)=W\left(-\frac{c}{b}e^{-ac/b}\right)$

$\displaystyle \displaystyle \frac{ac}{b}+cx\right=-W\left(-\frac{c}{b}e^{-ac/b}\right)$

$\displaystyle \displaystyle cx=-W\left(-\frac{c}{b}e^{-ac/b}\right)-\frac{ac}{b}$

$\displaystyle \displaystyle x=-\frac{1}{c}W\left(-\frac{c}{b}e^{-ac/b}\right)-\frac{a}{b}$