Let $a > 0$. I would like to know what is the largest index for the Fibonacci sequence $F\_n$ such that $F\_n \le a$. 

My effort: $F\_0 = 1$, $F\_1 = 2$, $F\_n = F\_{n-1} + F\_{n-2}$ and the explicit formula is $$F\_n={(1+\sqrt5)^{n+2}-(1-\sqrt5)^{n+2} \over 2^{n+2}\sqrt5}$$

So the inequation to solve is $F\_n\le a$. Can't find a way to solve this

I do know where $\phi$ = golden ratio that

$$F\_n = \left\lfloor {\phi^{n+2} \over \sqrt 5} \right\rfloor$$

this gives the result easily.

but I don't want to use this formula