A problem I'm working on requires me to solve $\ln r+\ln q=kr$ for $r$. I've tried using the Lambert $W$ function, but I'm not sure how to do it. Is there method, technique or known solution, that would help me approach this problem?
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1Set $u = kr$ and then you can rearrange for lambert.. But I am afraid (not totally sure) that $k$ and $r$ are forever tied. – Chinny84 Jun 10 '15 at 17:54
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Well, that or $q$. – Brian Tung Jun 10 '15 at 17:56
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But q is not equal to kr? – Nawar Ismail Jun 10 '15 at 17:58
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2There is no elementary way to isolate $r$, only the Lambert-function helps. – Peter Jun 10 '15 at 18:07
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Recall that
$$\log z=W(z)+\log W(z) \tag 1$$
where $W$ is Lambert's "W" function.
Here, we have
$$\log r+\log q=kr\implies -\log q=-kr+\log r \tag 2$$
So, let's add $\log (-k)$ to both sides of $(1)$ to obtain
$$\log(-k/q)=(-kr)+\log(-kr) \tag 3$$
Comparing $(3)$ with $(1)$ reveals that
$$r=-\frac1k\,W\left(-\frac{k}{q}\right)$$
NOTE:
If we are restricting $k$, $r$, and $q$ to be real-valued, then we must have $-\frac{k}{q}>-e^{-1}$.
Mark Viola
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My comment has been shot down! At least I put a "not totally sure" ;). +1 – Chinny84 Jun 10 '15 at 18:47
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@Chinny84 No worry. You were on the right track with $W$! And thank you for the up vote. Much appreciative. – Mark Viola Jun 10 '15 at 19:02