I have the functions $f(\lambda) = e^{\frac{\ln(\frac{-\lambda\theta^k}{3\lambda-2T})}{k}}$ and $g(\lambda) = \frac{\ln(\frac{-\lambda}{3\lambda-2T})}{k}+\theta$ where the constants $\theta, k, T$ are positive real numbers and $\lambda$ is a variable. I know that numerically these functions intersect at two points, $\lambda = 0.1$ and $\lambda \approx 0.113$ as indicated in the graph, but I would like derive a have a closed form expression for the intersection: 
I know that to find the intersection I must set $f(\lambda)= g(\lambda)$, but the algebra escapes me. Are there other methods we could use to find the intersection, or is messy algebra the only method?
Let $$\frac{\lambda }{2 T-3 \lambda }=x\implies f=\theta\, x^{\frac 1k}\quad \text{and}\quad g=\theta +\frac{\log (x)}{k}$$
Let $$y=x^{\frac 1k}\implies f=\theta\,y\quad \text{and}\quad g=\theta +\log(y)$$
So, you want to find the zeros of function $$F(y)=\theta\,y-\log(y)-\theta$$
The only explicit solution is given in terms of Lambert function $$y=-\frac{1}{\theta }W\left(-\theta\, e^{-\theta } \right)$$ and since the argument is negative, in the real domain, there are potentially two roots
$$y_1=-\frac{1}{\theta }W_0\left(-\theta\, e^{-\theta } \right)\qquad \text{and}\qquad y_2=-\frac{1}{\theta }W_{-1}\left(-\theta\, e^{-\theta } \right)$$
Notice that $y=1$ is a trivial solution. Then, the other root is $y_1$ or $y_2$.