TITLE: Leading Digit of $2^{4242}$
QUESTION [11 upvotes]: 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.

REPLY [2 votes]: 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 $1.6 \times 6 > 9$