Question about Benford's law A set of numbers is said to satisfy Benford's law if the leading digit d (d ∈ {1, ..., 9}) occurs with probability,
$$ P(d)=\log_{10}(d+1)-\log_{10}d$$
With Birkhoff's Ergodic Theorem  is possible to prove that the sequence $2^n$( for example) satisfies the Benford's law. 
The Fibonacci sequence satisfies Berford's law?
 A: Yes.If $x$ is a real value that is not a rational power of 10, then $x^n$ will satisfy Benford's law. To see this, note that $\log x$ is irrational, hence consider $\log x$ looping around the unit circle (fractional part). When $ \log d <  \{ n\log x \} < \log (d+1)$, then $x^n$ will have a starting digit of $d$. The sequence $\{ n \log x\}$ is uniformly dense on the unit circle, hence with probability $\log (d+1) - \log d$, $x^n$ will have a starting digit of $d$.
To apply this to the Fibonacci sequence, note that $F_n = A \phi^n + B \phi^{-n}$ and $\phi^{-1}< 1$ hence 'can be ignored'. So $\{ \log F_n \} \approx \{ \log A + n \log \phi \} $, and we have the result as needed.

The reason why we need $x$ to not be a rational power of 10, is so that $\log x$ is not a rational number. If it was, then $n \log x$ will be fixed points around the unit circle, and the starting digits will be extremely fixed. Whereas, when $\log x$ is irrational, it is standard to show that $\log x$ loops around the circle, doesn't return to the starting point, is periodic, hence has uniform density.
