If the number in question is known to be a power of 2, you are just trying to find $n$ such that $N=2^n$. If the input number is in decimal notation, just count the digits, subtract 1, multiply by 3.3219, and round up. Then add 1 if the initial digit is 2 or 3, 2 if the initial digit is 4, 5, 6, or 7, and 3 if the initial digit is 8 or 9.
For example, suppose $N=1267650600228229401496703205376$. This has 31 digits; $30\cdot3.3219 = 99.657$, so $N=2^{100}$. Or take $N=$
43699499387321412970609716695670835099367888141129535719972915195176
79444176163354392285807163181819981286546206512408458617685052043667
09906692902245553277900892247131030458103436298545516643924637451297
481464347472084863384057367177715867713536
which has 246 digits. $245\cdot3.3219 = 813.872383$; we round up to 814, add 2 because the first digit is 4, so this number is $2^{816}$.
The magic constant 3.3219 is actually $\log 10 / \log 2$. For input numbers in the hundreds of thousands of digits you will need a more accurate version, say 3.3219281.
The Fibonacci sequence represents a sort of "worse case" for the Euclidean algorithm. This occurs because, at each step, the algorithm can subtract $F_n\,\, $ only once from $F_{n+1} \,\,$. The result is that the number of steps needed to complete the algorithm is maximal with respect to the magnitude of the two initial numbers.
Applying the algorithm to two Fibonacci numbers $F {n}\,\, $ and $F_{n+1}\,\, $, the initial step is
$$ \gcd(F_{n},F_{n+1}) = \gcd(F_{n},F_{n+1}-F_{n}) = \gcd(F_{n-1},F_{n}) $$
The second step is
$$ \gcd(F_{n-1},F_{n}) = \gcd(F_{n-1},F_{n}-F_{n-1}) = \gcd(F_{n-2},F_{n-1}) $$
and so on. Proceding in this way, we need $n $ steps to arrive to $\gcd(F_{1},F_{2}) \,\,$ and to conclude that
$$ \gcd(F_{n},F_{n+1}) = \gcd(F_1,F_2) = 1$$
that is to say, two consecutive Fibonacci numbers are necessarily coprime.
Now it is well known that the growth rate of Fibonacci numbers, with respect to $n $, is exponential. In particular, $ F_n$ is asymptotic to
$${\displaystyle \varphi ^{n}/{\sqrt {5}}} $$
where
$$\varphi=\frac {1+\sqrt {5}}{2} \approx 1.61803$$
is the golden ratio. So, for $n $ sufficiently large, we have
$$n \approx \log_{\varphi} ( \sqrt {5}\,\, F_n) \\ = \log_{\varphi} ( F_n) + \frac {\log (5)}{2 \log (\varphi)} \approx \log_{\varphi} (F_n) $$
which tells us that the number of steps required to complete the Euclidean algorithm grows in a logarithmic manner with respect to $F_n\,\, $, and is bounded by the expression above. The same result can also be written using the logarithm in base $2$ as suggested by the OP:
$$n \approx \frac { \log_{2} F_n}{\log_{2} \varphi} =K \log_{2} F_n $$
where $$K=\frac {1}{\log_{2} \varphi}=\frac {\log {2}}{\log {\varphi}} \approx 1.4404... $$
Best Answer
One way is to divide the numerator and denominators by common factors (you do not need to compute the GCD explicitly). For example, $$\frac{5692}{84} = \frac{2\times2846}{2\times42} = \frac{2846}{42} = \frac{2\times1423}{2 \times 21} = \frac{1423}{21}.$$