As shown here, the (rounded) index $n$ of a given Fibonacci number $F$ is calculated with
$$ n(F) = \left\lfloor \log_\varphi (F \cdot \sqrt{5} + \frac12) \right\rfloor, $$
where
$$ \log_\varphi(x) = \frac{\ln(x)}{\ln(\varphi)} = \frac{\log_{10}(x)}{\log_{10}(\varphi)}. $$
Now my question is: How do I get there, e. g. from this formula?
$$ F_n = \left\lfloor \frac{\varphi^n}{\sqrt{5}} + \frac12 \right\rfloor, n \geq 0 $$
From Binet’s formula we have
$\sqrt{5}F_n=\phi^n-(-\phi)^{-n}$
and since $\phi^2 = \left(\frac{1+\sqrt5}{2}\right)^2 = \frac{1+2\sqrt5+5}{4} > \frac{10}{4} > 2$ we have
$\sqrt{5}F_n + \frac12 > \phi^n \space \forall n\ge 2$
But $\sqrt{5}F_n$ cannot be as great as $\phi^{n+1}$ so
$n<\log_{\phi}(\sqrt{5}F_n+\frac12)<n+1 \space \forall n \ge2$