I want to find the minimal number of bits for an RSA modulus giving a security, at least, of $2^{80}$.
For what I know the best algorithm to factorize an integer is GNFS (General Number Field Sieve in 1990 by A.Lenstra, H.Lenstra, Pollard, Manasse) which gives the result in subexponential time.
I've seen the following expression for the subexponential running time algorithms :
$$ L_{\alpha, c}(z) = O(e^{c.(ln z)^{\alpha}.(ln ln z)^{1-\alpha}}) $$ with $z$ an integer, $\alpha$ constant such $0 \leq \alpha \leq 1$, and $c$ a positive constant.
For GNFS, $c \approx 1,923$ and $\alpha = \frac{1}{3}$ , so I have to find $ln(z)$ or $z$ in : \begin{align} L_{[\frac{1}{3}, 1.923]} \geq 2^{80} \\ e^{1,92.(lnz)^{\frac{1}{3}}.(lnlnz)^{\frac{2}{3}}} \geq 2^{80} \end{align}
Once I have one of them, I have the solution doing $log_{2}(z)$.
The problem is I am stuck at the inequation, I can't solve it.
I tried : since the inequation's members are greater than $0$ and $ln$ is an increasing function, I can do : $$ln(e^{1,923.(lnz)^\frac{1}{3}.(lnlnz)^\frac{2}{3}}) \geq ln(2^{80}) $$ $$ 1,923.(lnz)^\frac{1}{3}.(lnlnz)^\frac{2}{3} \geq 80.ln(2)$$ $$ (lnz.(lnlnz)^{2})^{\frac{1}{3}} \geq \frac{80}{1,923}.ln(2)$$ I don't know if it's a good idea to do this way and how to end with $ln(z)$ or $z$.
I know the answer is 1024 bits but I want to prove it with an expression in z, I know it's easier by admitting 1024 bits length then prove the security is greater than $2^{80}$.