I'm sure I'm not the first one asking this but I couldn't find it in previous posts. We know that for any $\alpha>0$ we have $\lim_{x \rightarrow \infty} \frac{\ln x}{x^\alpha}=0$, but when looking at the graphs of these two functions for very small values of $\alpha$, it seems that $\ln x$ is larger for reasonable values of $x$.
So I'm trying to find $\tilde{x}$ such that $\forall x\geq \tilde{x}: x^ \alpha \geq \ln x$ (it would be best if $\tilde{x}$ was the minimal value such that this inequality holds, but that's not necessary).
I tried to solve for $x^ \alpha = \ln x$ (which by looking at the graphs I guess might have two roots?), and by substituting $t=x^\alpha$ we get $\ln t = \alpha t$ which is simpler, but I'm not sure if it has an explicit solution or some approximation.
I would appreciate any help or advice. Thank you!
As @Gary wrote in comments, the solution of $$x^\alpha=\log(x) \qquad \text{is}\qquad x=\left(-\frac{W(-a)}{a}\right)^{\frac{1}{a}}$$ A good approximation could be $$x=e \,\frac{1-\frac{1626 }{433}a+\frac{16751 }{4330}a^2-\frac{17089 }{17320} a^3} {1-\frac{2059 }{433}a+\frac{28681 }{4330}a^2-\frac{126439 }{51960}a^3 }$$ whose error is $\sim \frac{5a^7}{27}$
To give an idea of the accuracy $$\int_0^{\frac 9{10e}} \Bigg[\left(-\frac{W(-a)}{a}\right)^{\frac{1}{a}}-e \,\frac{1-\frac{1626 }{433}a+\frac{16751 }{4330}a^2-\frac{17089 }{17320} a^3} {1-\frac{2059 }{433}a+\frac{28681 }{4330}a^2-\frac{126439 }{51960}a^3 } \Bigg]^2\, da=3.01\times 10^{-5}$$
Edit
If you consider the second branch of Lambert function, let $u=-1-\log(a)$ and use the bounds given by Chatzigeorgiou in year $2013$ $$\log (a)- \sqrt{-2(\log (a)+1)} <W(-a)<\frac{1}{3} (2 \log (a)-1)- \sqrt{-2(\log (a)+1)}$$ to show that $x$ tends to infinity.
Just to give you an idea $$\left( \begin{array}{cc} a & x \\ 0.1000 & 3.43063\times 10^{15} \\ 0.0100 & 1.28544\times 10^{281} \\ 0.0010 & 7.94138\times 10^{3959} \\ 0.0001 & 4.31134\times 10^{50669} \\ \end{array} \right)$$