Decimal expansion of $2^n$ i am trying to read / understand the book: Introduction to the modern theory of dynamic systems, hassleblatt boris and katok anatole.
On page 28 there is an excercise which I am trying to solve but have no clue how it would work. I guess it is not very hard but I'm not getting any further.
I hope someone can help me.
1.3.1 Exercise:
Prove that the decimal expansion of the number $2^n$ may begin with any finite combination of digits.
 A: Here is a proof. Given any $a$, we want to argue the existence of $n$, $k$, and $r$ such that, $n,k,r \in \mathbb{N}$, $0 \leq r \leq 10^k - 1$, and
$$
2^n = a\times 10^k + r
$$
or equivalently, given any $a$ are there a $k$ and an $n$ such that,
$$
a\times 10^k \leq 2^n <(a+1)\times10^k,
$$
which can be written as
$$
\log(a)\leq n\log(2)-k<\log(a+1),
$$
where logarithms are base $10$. Now, we will use the following powerful theorem, known as Weyl's equidistribution theorem.
Theorem (Weyl's equidistribution theorem)
Let $\gamma$ be an irrational number, and $a,b$ be two real numbers such that $[a,b]\subset[0,1]$.  Then,
$$
\lim_{n\to\infty}\frac{1}{n}\left\{t : 1\leq t \leq n, t\in\mathbb{N},\{t\gamma\}\in[a,b]\right\} = (b-a)
$$
where $\{t\gamma\} = t\gamma - \lfloor t\gamma\rfloor$. 
To apply Weyl's theorem, let $l = \lfloor \log(a)\rfloor$. It is easy to observe that $[\log a - l,\log(a+1)-l]\subset[0,1]$ (except, I believe a trivial case, which is justified under the EDIT below), hence, applying Weyl's theorem to the interval $[\log a - l,\log(a+1)-l]$ with $\gamma = \log 2$ tell us that there must exists a $t$ such that $t\gamma - \lfloor t\gamma \rfloor$ is contained in the prescribed interval.
It suffices to pick, $\gamma = \log 2$ as above; $\mathbf{n=t}$, and $\mathbf{k=\lfloor t\gamma \rfloor - l}$ to finish the proof.
Remark : I had to rush a bit, so feel free to edit if there are missing, or non-rigorous pieces.
EDIT
The only case is the following: Still under $l \leq \log a < l+1$, if $\log(a+1)\geq l+1$. But in this case, since $a+1 \geq 10^{l+1}-1$, we must have
$$
10^{l+1}>a\geq 10^{l+1}-1
$$
and since $a$ is an integer, the only way to have this is $a = 10^{l+1}-1$. With this, the interval of interest becomes:
$$
[\log a ,\log(a+1)] = [\log a,l+1]
$$
and again after subtracting $l$, we end up in an interval between $[0,1]$. Hence, Weyl's theorem is valid.
A: We can take the logarithm of $2^x$ which gives $\log(2^x) = \frac{x\ln(2)}{\ln(10)}$ using the logarithm power rule and change of base formula. If we rewrite $\frac{\ln(2)}{\ln(10)}$ as $a$, it is evident that the equation $y = ax$ is a straight line since $a$ is constant.
Using the fact that $y = ax$ and $2^x$ are functions whose domains cover all real numbers, there exists a bijection (one-to-one correspondence) for each value of $2^x$ and $ax$. Thus if we choose $x$ accordingly, there will always be a value of $2^x$ that starts with any given sequence of numbers.
