Leading Digit of $2^{4242}$ How could I solve this problem?

Find the first digit of $2^{4242}$ without using a calculator.

I know how to find the last digit with modular arithmetic, but I can't use that here.
 A: If you have memorized that $\log_{10}2\approx 0.30103$ you can multiply by $4242$ and take the fractional part as $0.9692$ which looks like it should be greater than $\log_{10}9$ (and it is, but it is closer than I would have thought, it is $\approx 0.9542$).  I don't know how to do it without log tables.
A: Presenting an alternate method (no logs, but needs the knowledge that doubling time is ~ $70$/rate) 
Starting with 
$$2^{10}=1024$$ 
which is 
$$1000 \times 1.024$$ or a 2.4% increase. 
Then, 
$$70/2.4 \approx 29$$ 
implies that 
$$2^{290}\sim 2 \times 10^k$$ 
for some k. 
Then, $$2^{4242} = {2^{290}}^{14} \times 2^{182}$$
So, it would suffice to calculate $$2^{16} \approx 1.6 \times 10^l$$ 
and to get $2^{182}$, first note that $1.024^{29} \approx 2$ so, $1.024^{18} \approx 1.5$ (pure hand waving, but sounds logical) So, from that $$2^{182} \approx 6 \times 10^m$$ and thus we can get the fist digit to be close to $1.6 \times 6 > 9$
A: This is probably not the answer you are looking for, and wil probably only be appreciated by people of my age ...
I can still remember from school days that $\log_{10} 2 = 0,30102999$ (I always thought it was noteworthy that it is so close to $0.30103$) - people who went to school in the 1950's can probably recall using logs to base 10 for lots of tedious calculations.
You can then do the multiplication by 4242 without a calculator, and get the fractional part ($=x$, say, but you are likely to need a calculator to find out the first digit of $10^x$, unless you have also memorised $\log 2, \log3, \dots, \log 9$ (I can't!)
Edit:
With a bit more digging in the recesses of my memory, I can just recall that $\log 3$ is something like $0.477$, so $\log 9 = 2 \log 3 = 0.954$, so that should do it ...
