$\ln r+\ln q=kr$ Isolating $r$

Question

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?

Answer 1

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

$$r=-\frac1k\,W\left(-\frac{k}{q}\right)$$

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NOTE:

If we are restricting $k$, $r$, and $q$ to be real-valued, then we must have $-\frac{k}{q}>-e^{-1}$.