Inspired by Kinma’s brilliant post on calculating 10 to decimal powers https://forum.artofmemory.com/t/calculating-10-0-1-0-2-0-3-etc/29957
I’d like to do the same for 2. This is useful for measuring how much something has grown, doubled etc. I know that 2^0.5 is 1.41 but can’t get much beyond that. What are the next steps in calculating:
2^0.1
2^0.2
2^0.3
2^0.4
2^0.6
2^0.7
2^0.8
2^0.9
would like to also then advance to 2^1.1 etc
The trick I used for 10^{0.1} is based on the fact that 10^{0.301} \approx 2, then use 0.3 instead of 0.301 and later make a correction to make up for the difference between 0.3 and 0.301.
I have been thinking if I could come up with something similar for powers of 2. However I could not find anything.
Not to worry; we already have a simple rule to work out 2^{0.1}. The rule of 72 of course.
Using this rule we find that 2^{0.1} = 1.072.
I leave the rest of the series as an exercise to you.
I can offer 2^0.5 = 1.41, the rest are a little harder. Using index laws a^b x a^c = a^(b+c) we can do 2^0.6 = 2^0.5 x 2^0.1 = 1.41 x 1.07 = 1.51 which I can just barely do in my head. I guess that’s why we try and use 10 wherever possible in Mental arithmetic, it makes every operation easier. This I believe also goes for indexes and logarithms, base 10 is much easier than other log bases.
Interestingly as I delve through logs I see that (10^0.3)^1/3 is the same as 2^1/3 again utilising Kinma’s axiom that 10^0.3 is approximately equal to 2. Possibly then, to complete this exercise we need to look at utilising the log 10 base conversion to obtain our missing powers of 2.