Logarithms come in handy when searching for power laws. Suppose you have some data points given as pairs of numbers $(x,y)$. You could plot a graph directly of the two quantities, but you could also try taking logarithms of both variables. If there is a power law relationship between $y$ and $x$ like
$$y=a x^n$$
then taking the log turns it into a linear relationship:
$$\log(y) = n \log(x) + \log(a)$$
Finding the exponent $n$ of the power law is now a piece of cake, since it corresponds to the slope of the graph.
If the data do not follow a power law, but an exponential law or a logarithmic law, taking the log of only one of the variables will also reveal this. Say for an exponential law
$$y=a e^{b x}$$
taking the log of both sides gives
$$\log(y) = b x + \log(a)$$
Which means that there will be a linear relationship between $x$ and $\log(y)$.
To address some preliminaries, let $T(a,b)$ be the number of steps taken in the Euclidean algorithm, which repeatedly evaluates $\gcd(a,b)=\gcd(b,a\bmod b)$ until $b=0$, assuming $a\geq b$. Also, let $h=\log_{10}b$ be the number of digits in $b$ (give or take). (Note that in these calculations, by counting steps, we ignore the question of the time-complexity of the $\mathrm{mod}$ function. If we assume it is $O(1)$, then all of the following also applies to the time complexity of the algorithm.
In the worst-case, as you have stated, $a=F_{n+1}$ and $b=F_n$, where $F_n$ is the Fibonacci sequence, since it will calculate $\gcd(F_{n+1},F_n)=\gcd(F_n,F_{n-1})$ until it gets to $n=0$, so $T(F_{n+1},F_n)=\Theta(n)$ and $T(a,F_n)=O(n)$. Since $F_n=\Theta(\varphi^n)$, this implies that $T(a,b)=O(\log_\varphi b)$. Note that $h\approx log_{10}b$ and $\log_bx={\log x\over\log b}$ implies $\log_bx=O(\log x)$ for any $a$, so the worst case for Euclid's algorithm is $O(\log_\varphi b)=O(h)=O(\log b)$.
The average case requires a bit more care, as it depends on the probabilistics of the situation. In order to precisely calculate it, we need a probability distribution. If $a$ is fixed and $b$ is chosen uniformly from $\mathbb Z\cap[0,a)$, then the number of steps $T(a)$ is
$$T(a)=-\frac12+6\frac{\log2}\pi(4\gamma-24\pi^2\zeta'(2)+3\log2-2)+{12\over\pi^2}\log2\log a+O(a^{-1/6+\epsilon}),$$
or, for less accuracy, $T(a)={12\over\pi^2}\log2\log a+O(1)$. (Source: Wikipedia)
In the best case, $a=b$ or $b=0$ or some other convenient case like that happens, so the algorithm terminates in a single step. Thus, $T(a,a)=O(1)$.
If we are working on a computer using 32-bit or 64-bit calculations, as is common, then the individual $\bmod$ operations are in fact constant-time, so these bounds are correct. If, however, we are doing arbitrary-precision calculations, then in order to estimate the actual time complexity of the algorithm, we need to use that $\bmod$ has time complexity $O(\log a\log b)$. In this case, all of the "work" is done in the first step, and the rest of the computation is also $O(\log a\log b)$, so the total is $O(\log a\log b)$. I want to stress, though, that this only applies if the number is that big that you need arbitrary-precision to calculate it.
(This underscores the difference between the mathematician's big-O and the programmer's big-O notation: in the first case, you want the bound to be true $\forall n$, even those $n$ which are so absurdly large that no one can ever write them down or store them in memory, whereas in the second case, programmers are primarily interested in $n\in[0,2^{64}]$, and that's a liberal estimate. For them, it's more important to see the "leading contribution" to the time complexity, and for the Euclidean algorithm, the smaller number drives the difficulty of the calculation by and large.)
Best Answer
Suppose that $f$ is $O(\log(\log n))$, and $g$ is $\Theta(\log n)$. Then there are positive constants $c_0$ and $c_1$ and an $m\in\Bbb Z^+$ such that
$$f(n)\le c_0\log(\log n)$$
whenever $n\ge m$ and
$$g(n)\ge c_1\log n$$
whenever $n\ge m$. Moreover,
$$\lim_{n\to\infty}\frac{\log(\log n)}{\log n}=0\;,$$
so we may further assume that $m$ is large enough so that
$$\frac{\log(\log n)}{\log n}\le\frac{c_1}{c_0}$$
whenever $n\ge m$.
But then for $n\ge m$ we have
$$f(n)\le c_0\log(\log n)\le c_1\log n\le g(n)\;,$$
i.e., $g$ is eventually an upper bound for $f$.