So i have an upcoming exam, and since no calculators is allowed, i was wondering if there is an approach to calculating the value of 2 to the power of some value?
For example,
2^4,2^5, 2^12, 2^13.
I know a way is as such where:
2^4 = 2x2x2x2 = 16
2^5 = 2x2x2x2x2 = 32
2^12 = 2x2x2x2x2x2x2x2x2x2x2x2 = 4096
But it would definitely take a long time for power of 12 and 13, so is there any faster approach on solving these without calculator?
On
Memorization and repeated squaring will get you there. If you want $2^{12}$, you can write it as $((2^3)^2)^2$ and then $2^3=8, 8^2=64, 64^2=4096$ Presumably you know the first two, so you only need to square $64$. If you want $2^{19}$ you can write it as $2^{16} \cdot 2^3=65536\cdot 8$. If you have a test like this (which I abhor) you should know the powers of $2$ up to at least $2^{16}$
On
You can get to to $2^{10}=1024$ by doubling and squaring and you can get $2^{20}$ by knowing $2^{10} =1000 + 25 -1 = 100 + \frac {100}4 - 1$ and distributing.
So for example: $2^{17} = 2^{10}2^7 = (1000 + \frac {100}4 - 1)*2^7= 1000*2^7 + 100*2^5 - 2^7$.
$2^3= 8$ so $2^7 =8*8*2 = 64*2=128$. And $2^5 = 2^4*2 = 16*2=32$.
So $2^{17} = 128,000 + 3,200 - 128 = 131,100 - 28 = 131,070 + 2 = 131,072$
We can take it further $2^{20} = (1000 + 24)^2 = 1,000,000 + 48,000 + (25-1)^2 = 1,048,000 + 625 -50 + 1 = 1,048,576$. This gets iffy.
$2^{20} = 1,000,000 + \frac{100,000}2 - \frac{10,000}4 + 1,000 + \frac {100}2 + \frac {100}4 + 1$ so $2^{26}$ say is $64,000,000 + 3,200,000 - 160,000 + 64,000 + 3,200 + 1,600 + 64 = 67,108,864$.
It's not easy but it becomes "familiar looking"
You know $2^3 =8$ so $2^6 = 8^2=64$ and $2^8=16^2=256$, and $2^{10} = 4\times 256=1024$, $2^{13}=8192$,....