Consider the equation
$$K=\frac{gR_{0}\left(r+1\right)^{t}\left(\left(w+1\right)^{N}-\left(r+1\right)^{N}\right)}{\ln\left(w+1\right)-\ln\left(r+1\right)},$$
such that $g,R_0,r,w\in \mathbb{R}$ and $N\in\mathbb{R}^+$. Can this be solved for $r$? Any help would be much appreciated.
For background, the above is a solution to the differential equation
$$S\cdot\frac{dP}{dt}=\frac{I\left(t+N\right)}{S\left(1+w\right)^{t+N}}-\frac{I\left(t\right)}{S\left(1+w\right)^{t}}$$
where $I(t):=\left(1-\tau\right)\left(gR_0(1+r)^t-J\right)$ and the constant $K=\frac{SP+C}{\left(1-T\right)\left(i+1\right)^{-N-t}}-\frac{J\left(1-\left(i+1\right)^{N}\right)}{\ln\left(i+1\right)}$.
Let $$x=\left(\frac{r+1}{w+1}\right)^{\frac{1}{t}} \qquad \text{and} \qquad A=\frac{K }{g R_0 t(w+1)^{N+t}}\qquad \text{and} \qquad a=\frac N t$$ and the equation becomes $$A \log(x)= x(x^a-1)$$ If $x\gg 1$, then we have (as an approximation) $$x\sim \Bigg[-\frac{A }{1+a}W_{-1}\left(-\frac{1+a}{A}\right)\Bigg]^{\frac 1{1+a}}$$ from which you could start Newton method.
Suppose $t=N=A=10$ , this would give $x=3.56556$ and Newton iterates will be
$$\left( \begin{array}{cc} n & x_n \\ 0 & 3.56556 \\ 1 & 4.63742 \\ 2 & 4.38790 \\ 3 & 4.37384 \\ 4 & 4.37380 \end{array} \right)$$
For the case where $t=0$, let
$$x=\left(\frac{r+1}{w+1}\right)^{\frac{1}{N}} \qquad \text{and} \qquad A=\frac K{g R_0 N(w+1)^N}$$ which makes $$A \log(x)=x-1 \implies x=-A\, W_{-1}\left(-\frac{1}{A}e^{-1/A}\right)$$