Consider the powers of $2$:

They always end in $2$, $4$, $6$ or $8$, but we know little about their beginnings

1. Is there a power of $2$ starting with $9$?
2. Is there a power of $2$ starting with $9999\cdots 9999$ ($99$ nines)?
3. When considering more and more powers of $2$, which is the average of powers starting with $9$ in the limit?

1. There are powers of $2$ starting with $9$, for instance $2^{53} = 9007199254740992$
2. Now we can't proceed by direct inspection, because humongous numbers are involved... The right direction to prove that there are powers of $2$ starting with $9999\cdots 9999$ ($99$ nines) is to work with logarithms to the base $10$ (denoted $\log$): if some power starts with $9999\cdots 9999$

$$2^n=9999\cdots 9999\cdots\cdots\cdots$$

then

$$n\log 2=\log(2^n)=\log(9999\cdots 9999\cdots\cdots\cdots)=\log(9.999\cdots 9999\cdots\cdots\cdots)+m$$

where $m$ is the number of digits of $9999\cdots 9999\cdots\cdots\cdots$ minus $1$. Considering the fractional part,

$$\{n\log 2\}=\log(9.999\cdots\cdots\cdots)$$

that is, we're looking for some $n$ that satisfies

$$\log(9.999\cdots 9999)\leqslant\{n\log 2\}\lt\log(10)=1$$

and this is possible, because $\log 2$ is an irrational number (check it out! $2^q\neq 10^p$ for positive integers $p$ and $q$)! And this ensures that the fractional parts of $n\log 2$ are dense in the interval [0,1]: one may find arbitrarily close elements of this set to any given number $0\leqslant x\leqslant 1$. So we only have to wait for some $\{n\log 2\}$ to fall into $[\log(9.999\cdots 9999),1)$; we will have to wait a lot, sure, but there will be such powers

And with the same argument, we find out that there are powers of $2$ starting with any given finite sequence of numbers. Cool, isn't it?

3. Since in the limit the values $\{n\log 2\}$ are equally distributed in the interval $[0,1]$, the average of powers starting with $9$ is

$$\log 10-\log 9\simeq 0.0458$$

That's quite interesting: there are powers of $2$ starting with any number, but bigger numbers appear less. Had you ever thought about it?