Trying to answer this question where we look for the solution of $$\large\color{red}{t^k+t=1} \qquad \qquad \text{with} \qquad \color{red}{0<k<1}$$ which is more or less the function Lambert considered.
In my update, I rewrote it as $$\large\color{blue}{k=\frac{\log (1-t)}{\log (t)}}$$
The plot of $t$ as a function $k$ is not very nice but my surprise came from the plot of $t$ as a function of $\log(k)$ which is extremely close to a logistic function.
I came very quickly to the approximate result $$\large\color{blue}{t\sim\frac 1 {1+k^{-\log_2 (\phi )}}}$$ ($\phi$ being the golden ratio). This reproduces exactly the value of $t$ for $k=\frac 12$; at this point, the slope is in a relative error of $0.3$%.
This surprising value being used as the $t_0$ of Newton method, some results
$$\left( \begin{array}{cccc} k & t_0 & t_1 & \text{solution} \\ 0.05 & 0.11107938 & 0.10607326 & 0.10610459 \\ 0.10 & 0.16818422 & 0.16491137 & 0.16492096 \\ 0.15 & 0.21130781 & 0.20917626 & 0.20917956 \\ 0.20 & 0.24650516 & 0.24512115 & 0.24512233 \\ 0.25 & 0.27639320 & 0.27550762 & 0.27550804 \\ 0.30 & 0.30240988 & 0.30186067 & 0.30186067 \\ 0.35 & 0.32545141 & 0.32513010 & 0.32513010 \\ 0.40 & 0.34612246 & 0.34595481 & 0.34595481 \\ 0.45 & 0.36485403 & 0.36478837 & 0.36478837 \\ 0.50 & 0.38196601 & 0.38196601 & 0.38196601 \\ 0.55 & 0.39770340 & 0.39774335 & 0.39774335 \\ 0.60 & 0.41225852 & 0.41232020 & 0.41232020 \\ 0.65 & 0.42578544 & 0.42585602 & 0.42585602 \\ 0.70 & 0.43840971 & 0.43848026 & 0.43848026 \\ 0.75 & 0.45023507 & 0.45029952 & 0.45029952 \\ 0.80 & 0.46134837 & 0.46140274 & 0.46140274 \\ 0.85 & 0.47182302 & 0.47186491 & 0.47186491 \\ 0.90 & 0.48172173 & 0.48174985 & 0.48174985 \\ 0.95 & 0.49109845 & 0.49111241 & 0.49111241 \\ \end{array} \right)$$
To give an idea, I considered as a measure $$\Phi_n=\int_0^1 \Bigg[k-\frac{\log (1-t_n)}{\log (t_n)}\Bigg]^2\,dk$$ $$\Phi_0=2.157 \times 10^{-6}\qquad \Phi_1=6.282 \times 10^{-11}\qquad \Phi_2=4.007 \times 10^{-18}$$
Edit
After @Jam's answer, I minimized $$\Psi(a)=\int_0^1 \Bigg[k-\frac{\log \left(1+k^a\right)}{\log \left(1+k^{-a}\right)} \Bigg]^2\,dk$$ The result is $$a_{\text{min}}=0.69603517 \quad \implies \quad \Psi(a_{\text{min}})=1.668 \times 10^{-6}$$
For this number, the $ISC$ proposes the amazing $$a_{\text{min}}\sim \frac{\sqrt[2]{2}\,\, \sqrt[3]{3}-9}{10} $$
Could this be even partly justified ?