Leading Digit of $2^{4242}$

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 ...

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.

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 $1.6 \times 6 > 9$