# 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.

Relevance
• 2 months ago

The simplest is using logarithms.

Using log to the base 10

n^m = 10^(log(n^m)) = 10^(m*log(n))

so:

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.

or:

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

so n^m

= 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.

• 2 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

• ?
Lv 7
2 months ago

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?

• Dixon
Lv 7
2 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.

• ?
Lv 7
2 months ago

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.

• 2 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