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Is there a method to raise a number to a power other than by repeated multiplication?
Is there an algorithm, short cut, or trick to raising an integer to a large integer power other than by multiplying by itself over and over?
I'm asking for a pencil/paper calculation, not a computer program.
- L. E. GantLv 72 months ago
The simplest is using logarithms.
Using log to the base 10
n^m = 10^(log(n^m)) = 10^(m*log(n))
look up log(n)
multiply by m
use antilog (...) to get n^m
or, use a calculator -- your computer software most likely includes a calculator.
the power is either odd or even
if even, then it's of the form 2k
if odd, then it has the form 2k+1
= n^(2k) or n^(2k+1)
= (n^k)^2 or (n^k)^2 * n
You can apply the same technique at every level down to n^1
Using 2 as n...
2^10 = 2^(2*5) = 4^5
4^5 = 4^(2*2 + 1) = 16^2 * 16 etc.
number of iterations is far less than you'd expect.
- Demiurge42Lv 72 months ago
One of the easiest methods is to square the base then keep squaring your result until you get a power that's greater than what you are looking for.
2^23 for example.
Instead of multiplying 22 times, calculate the following
2^1 = 2
2^2 = 4
2^4 = 4^2 = 16
2^8 = 16^2 = 256
2^16 = 256^2 = 65,536
The next square would be 2^32 which is bigger than what you are trying to find.
23 = 16 + 4 + 2 + 1 so now you can use the previous results and the properties of exponents to get:
2^23 = (2^16)(2^4)(2^2)(2^1) = (65,536)(16)(4)(2) = 8,388,608
A better question is why would you want to use logs when a calculator would give you results like 2^123 = 10633823966279326983230456482242756608
in just a few seconds?
- DixonLv 72 months ago
There are a few tricks for squaring certain numbers and things like that but generally speaking, if you want a value you just have to do repeated multiplication.
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It would help to have either a slide rule or a table of logarithms. Years ago, when we had no calculators, I did like to use a one-page table of logarithms.
Even if you know only a FEW common logarithms in your head, they can be helpful. For instance, I know log(2) = 0.301, log(3) = 0.477, log 7 = 0.845.
From that I can deduce that log(5) = 0.699 and log(6) = 0.778.
Now let's say I want to raise 65 to the 65th power. Using the facts just stated, I can see that log(6.5) is probably around 0.81 and so log(65) is around 1.81. Another approach would be to note that 64 = 2^6 and therefore log(65) will be be a little over 6*log(2) = 1.806. Anyhow, if log(65) = 1.81, then
log(65 ^ 65) = 65*log(65) = (enter pencil and paper) = 117.65, which is
log(10^117) + 0.65. And 0.65 might be near log(5), refer to paragraph above.
So I know that 65 to the 65th power is a 118-digit number and that the first digit might be a 4 or 5.
It turns out this isn't a great estimate. 65^65 is indeed a 118-digit number, but the first two digits are 6.9.
HOWEVER, if I had had a log table in hand, I'd have looked up log(6.5) and found that it's 0.813, so log(65) = 1.813, and 65*log(65) = (pencil and paper) = 117.845, so I'd have said 65^65 is a 118-digit number whose first digit is 7. And that would've been a very good estimate !
If this is hard to follow, get hold of an OLD algebra book and read up on common logarithms.
- billrussell42Lv 72 months ago
no, that is about it, unless you can use log tables.
with log tables, 5^7, for example..
log 5^7 = 7 log 5 = 4.89279
now use anti-log table to get 78125
or a good sliderule