Solving inequality with Lambert W

Question

I'm new to the Lambert W function and need to solve an inequality in the following form for n:

$bcn-d\cdot (\frac{1}{a})\cdot log(1+axn)>0$

All parameters b,c,d,a and x are positive and real. Appreciate any help very much!

Answer 1

If you have this inequality, it first means that you need to solve for $n$ the equation $$b c n-\frac{d }{a}\log (1+a n x)=0 \tag 1$$

I do not know why the Wipkipedia page changed quite recently. It used to contain a series of good and simple examples showing the steps to follow in order to arrive to something looking like $z e^z=k$ or its equivalent taking logarithms. Just trying to do it, start with $1+a n x=z$, that is to say $n=\frac{z-1}{a x}$ provided that $ax \neq 0$. So, the equation becomes $$b c z- dx \log(z)= b c\implies bcz-dx \log(bcz)=bc-dx \log(bc)$$ Now $bcz=u$ and just continue.

After all the required substitution steps, back to $n$, you should arrive (if I am not mistaken) to $$n=-\frac{1}{a x}-\frac{d }{a b c}W\left(-\frac{b c }{d x}e^{-\frac{b c}{d x}}\right)$$ Take care that in the real domain $W(t)$ exists only if $t \geq -\frac 1e$ and that for $-\frac 1e \leq t \leq 0$, there is the other branch $W_{-1}(t)$ which is real.

Edit

Go here for the examples.