According to Wikipedia, it is possible to invert Binet's formula for Fibonacci numbers:
$$F_n = \frac{\varphi^n-\psi^n}{\varphi-\psi} = \frac{\varphi^n-\psi^n}{\sqrt 5}$$ where $\varphi = \frac{1 + \sqrt{5}}{2} \approx 1.61803\,39887\ldots$ and $\psi = \frac{1 - \sqrt{5}}{2} = 1 - \varphi = - {1 \over \varphi} \approx -0.61803\,39887\ldots$, or more specifically the truncating variant of that:
$$F_n = \left\lfloor\frac{\varphi^n}{\sqrt 5} + \frac{1}{2}\right\rfloor$$
to find the index $n(F)$ of the largest Fibonacci number that is not greater than a real number $F > 1$:
$$n(F) = \left\lfloor \log_\varphi (F\sqrt5 + 1/2)\right\rfloor$$
Since no justification is given, besides the floor function being monotonic, I would like to prove this.
So assuming the truncating formula holds: $$F_n = \left\lfloor\frac{\varphi^n}{\sqrt 5} + \frac{1}{2}\right\rfloor$$ $$F_n = \frac{\varphi^n}{\sqrt 5} + \frac{1}{2} + E, \text{where $0 \le E < 1$}$$ $$\varphi^n = \sqrt5(F_n - \frac{1}{2} - E), \text{where $0 \le E < 1$}$$ $$n \log\varphi = \log(\sqrt5(F_n - \frac{1}{2} - E)), \text{where $0 \le E < 1$}$$ $$n = \log_\varphi(\sqrt5(F_n - \frac{1}{2} - E)), \text{where $0 \le E < 1$}$$
Since n is a whole number we can take the floor:
$$n = \lfloor n \rfloor = \left\lfloor \log_\varphi(\sqrt5(F_n - \frac{1}{2} - E)) \right\rfloor, \text{where $0 \le E < 1$}$$ $$n = \left\lfloor \log_\varphi(\sqrt5 F_n \left(1 - \frac{1 + 2E}{2F_n}\right) \right\rfloor, \text{where $0 \le E < 1$}$$ $$n = \left\lfloor \log_\varphi(\sqrt5 F_n) + \log_\varphi \left(1 - \frac{1 + 2E}{2F_n}\right) \right\rfloor, \text{where $0 \le E < 1$}$$
And this does not seem to be the right path...
