I need to solve the next equation x:
$d-x+yln[\frac{d}{x}]=b$
I inserted this into Wolfram Alpha and it returned:
$x = y \Bbb{W}[\frac{e^\frac{d-b}{y}d}{y})]$
y, d, b, and x are all real, positive numbers.
How do I solve for x by implementing the lambert W function?
Thanks!
Not sure if this'll help. You've probably read by now that the Lambert W-Function is defined as the inverse relation of $xe^x$. That is to say, $W(xe^x)=x$.
You can usually just leave the $W$ function as it is, or make a numerical computation using WolframAlpha, or various numerical methods (Newton's method, Halley's method, etc.) to approximate it.
So, if you want to plug in $d,b,y$ values to find the various $x$-values, then you'll just need to use numerical methods to solve it (see @robjohn's answer).
There is no way to further simplify. The reason the function even exists, is because it is a special function which has been defined for this purpose. You can't express it in terms of elementary functions.
Now, to get the Wolfram result:
$d-x+y\ln[\frac{d}{x}]=b$
$d-x+y[\ln d-\ln x] =b$
$x+y\ln x =y\ln d+d-b$
$m=y\ln d+d-b$
$x+y\ln x=m$
$\frac{x}{y}+\ln x=\frac{m}{y}$
$xe^{\frac{x}{y}}=e^{\frac{m}{y}}$
$\frac{x}{y} e^{\frac{x}{y}} = \frac{1}{y} e^{\frac{m}{y}}$
Taking the product log of both sides, we end up with:
$\frac{x}{y} = W[\frac{1}{y} e^{\frac{m}{y}}]$
So, solving for $x$, we get that:
$x = y W[\frac{1}{y} e^{\frac{m}{y}}]$
To get the result in the form of Wolfram's answer, just substitute back in for $m$:
$x = y W[\frac{1}{y} e^{\frac{y\ln d+d-b}{y}}]$
$x = y W[\frac{1}{y} e^{\ln d+\frac{d-b}{y}}]$
$x = y W[\frac{1}{y} d e^{\frac{d-b}{y}}]$
And that's as far as we can get. If you want to numerically evaluate $x$, then you can either use iterative processes, such as are on the Wikipedia page, or just use Wolfram to calculate.
This answer gives the following method for computing Lambert W:
Analysis of $we^w$
For $w\gt0$, $we^w$ increases monotonically from $0$ to $\infty$. When $w\lt0$, $we^w$ is negative.
Thus, for $x\gt0$, $\mathrm{W}(x)$ is positive and well-defined and increases monotonically.
For $w\lt0$, $we^w$ reaches a minimum of $-1/e$ at $w=-1$. On $(-1,0)$, $w e^w$ increases monotonically from $-1/e$ to $0$. On $(-\infty,-1)$, $w e^w$ decreases monotonically from $0$ to $-1/e$. Thus, on $(-1/e,0)$, $\mathrm{W}(x)$ can have one of two values, one in $(-1,0)$ and another in $(-\infty,-1)$. The value in $(-1,0)$ is called the principal value.
The iteration
Using Newton's method to solve $we^w$ yields the following iterative step for finding $\mathrm{W}(x)$: $$ w_{\text{new}}=\frac{xe^{-w}+w^2}{w+1} $$
Initial values of $w$
For the principal value, when $-1/e\le x\lt0$, and when $0\le x\le10$, use $w=0$. When $x\gt10$, use $w=\log(x)-\log(\log(x))$.
For the non-principal value, when $x$ is in $[-1/e,-.1]$, use $w=-2$; and if $x$ is in $(-.1,0)$, use $w=\log(-x)-\log(-\log(-x))$.
