Is $v-v_\circ=\beta\ln\left(\frac{\beta+v}{\beta+v_\circ}\right)$ solvable for $v$?

Question

In the equation

$$v-v_\circ=\beta\ln\left(\frac{\beta+v}{\beta+v_\circ}\right)$$

where $v_\circ\,, \beta \gt0$ and $v \in (-\beta , v_\circ]$

Turns out that we are unable to represent $v$ as a combination of elementary functions of $\beta$ and $v_\circ$.

And the Lambert's W function is helpful here.

Then, how can we solve the above equation?

Answer 1

$$v-v_\circ=\beta\ln\left(\frac{\beta+v}{\beta+v_\circ}\right)$$ $$e^{v-v_\circ}=e^\beta\left(\frac{\beta+v}{\beta+v_\circ}\right)$$ $$(\beta+v_\circ)e^ve^{-\beta-v_\circ}=\beta+v$$ $$-(\beta+v_\circ)e^{-2\beta-v_\circ}=(-\beta-v)e^{-\beta-v}$$ $$-\beta-v=W_k(-(\beta+v_\circ)e^{-2\beta-v_\circ})$$ $$v=-\beta -W_k(-(\beta+v_\circ)e^{-2\beta-v_\circ})$$ Where $W_k$ is the lambert W function.

Answer 2

Further to Wong Austin's answer, the multi-valued nature of complex exponentiation gives the Lambert W multiple branches. However, one root is obvious. The equation can be rearranged as $f(v)=f(v_\circ)$ with $f:=v-\beta \ln(\beta+v)$. Thus $v=v_\circ$ works.