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This question already has an answer here:

Given that the Fibonacci number is 165580141 Is it possible to find $n$ using the closed form. I tried simplifying the closed form but I get stuck at:

$\sqrt5F_n = (\frac{(1+\sqrt5}{2})^n - (\frac{(1+\sqrt5}{2})^n$

$\sqrt5(165580141) + (\frac{(1+\sqrt5}{2})^n = (\frac{(1+\sqrt5}{2})^n$

($370248451 + (\frac{1+\sqrt5}{2})^n)^{1/n} = \frac{(1+\sqrt5)}{2}$

I don't know how to proceed further.

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marked as duplicate by MJD, Leucippus, user99914, Shailesh, AugSB Sep 27 '17 at 7:33

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    $\begingroup$ This is a very small number. $n=41$ if you start from $F_1=F_2=1$. $\endgroup$ – lulu Sep 26 '17 at 21:24
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You mean $$\sqrt5 F_n=\left(\frac{1+\sqrt5}2\right)^n-\left(\frac{1-\sqrt5}2\right)^n.$$ The second term is very small so $$\sqrt5 F_n\approx\left(\frac{1+\sqrt5}2\right)^n.$$ If you know $F_n$ and want to find $n$, why not take logarithms?

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Just for illustration of Lord Shark the Unknown's suggestion.

Define $k_n$ as $$k_n=\frac{\log \left(\sqrt{5} F_n\right)}{\log (\phi )}$$ and compute for a few values of $n$ $$\left( \begin{array}{cc} n & k_n \\ 3 & 3.112696029 \\ 4 & 3.955287767 \\ 5 & 5.016827815 \\ 6 & 5.993536209 \\ 7 & 7.002463652 \\ 8 & 7.999058197 \\ 9 & 9.000359624 \\ 10 & 9.999862619 \\ 11 & 11.00005247 \\ 12 & 11.99997996 \\ 13 & 13.00000766 \\ 14 & 13.99999708 \\ 15 & 15.00000112 \\ 16 & 15.99999957 \\ 17 & 17.00000016 \\ 18 & 17.99999994 \\ 19 & 19.00000002 \\ 20 & 19.99999999 \end{array} \right)$$

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