I've been reading about Exponents, and I was wondering if there is a shorter way to do this same calculation, below:

24 = 2 * 2 * 2 * 2 = 16

I keep seeing what seems to me a pattern in this, and other examples. The pattern I keep seeing (maybe it's a coincidence?), is that they always seem to only multiply the base number by the base number, until it reaches the same value that you would reach if you just multiplied the exponent by itself, just once.

What I mean is, can we not just do it this way instead? 24 = 4 * 4 = 16

I.e. Instead of multiplying the base by the base a billion times, we just multiply the exponent by itself, once.

And how do we calculate the power of using just the Windows calculator?

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A rule which only works in one (or just a handful) of cases is not a very good rule to use. –  rschwieb Feb 27 '13 at 14:41
I agree. But I have no idea about exponents so I thought I would ask. –  Tommy Feb 28 '13 at 9:59
Yes, but what I'm getting at is that if you have a conjecture based on one observation only, you might try a few other random combinations to see if it holds, rather than proceeding with the conjecture :) –  rschwieb Feb 28 '13 at 13:59

The pattern is a only coincidence that occurs when the base is 2. Take an example when the base is 3:

$3^4 = 3*3*3*3 = 81$

$3^4$ is not $4*4 = 16$

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Thank you very much, @Cheeku! So, in every case, all that you need to do, is to multiply the base by the number of the exponent. But what if the exponent is a large number, like 500? Is there a shorter way to calculate it? –  Tommy Feb 27 '13 at 14:17
You approximate them with something known as the concept of logarithms and anti-logarithms –  Cheeku Feb 27 '13 at 14:19
Thank you. I will look them up. I will accept the answer when the timer runs out. –  Tommy Feb 27 '13 at 14:20
If the exponent is a large number, as in $2^{500}$, you can save a lot of multiplication by calculating $2^{250}$ and then multiplying that once by itself. To calculate $2^{250}$ you can calculate $2^{125}$ and multiply it once by itself. To calculate $2^{125}$ you can calculate $2^{62}$, multiply it by itself to get $2^{124}$ and the multiply by 2 to get $2^{125}$. Doing it this way takes a lot less arithmetic than multiplying by two 499 times. –  MJD Feb 27 '13 at 15:00

What you saw is a bit of a coincidence since $2^2=4$.

What was used here is this: $$2^{4}=2^{2\cdot2}=(2^{2})^{2}=4^{2}=16$$

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Your result is a coincidence. The reason it works for the example is,$2^4=(2^{2.2})=(2^2)^2=4^2$ It is because $a^{bc}=(a^b)^c$. If speed in exponentiation is what you want, the standard divide and conquer algorithm will work. It goes like this: $a^{2b}=(a^b)^2$, $a^{2b+1}=a(a^b)^2$

For example

$2^{10}=(2^5)^2$

$2^5=2(2^2)^2$

$2^2=4$

$2^5=2.4.4=32$

$2^10=32.32=1024$

Note, that this reduces the no. of multiplications. It may not be of much help to you as it is already implemented in all calculators, and is tedious to do by hand.

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You certainly all know about the coincidence $$\frac{\not \!6 4}{1 \!\!\not \!6} = 4.$$