Expert: Scott A Wilson - 6/20/2011

Question
When it comes to 5^5 I understand that this is a short way of saying 1X5X5X5X5X5. I understand that 5^-5 is a short way of saying 1/5/5/5/5/5. However, I do not understand what to think of 5^.5

I realize that this is another way to say the square root of five, but I do not understand why. This question came up as I was studying logarithms and the number 'e'. Any help would be appreciated,

Aaron

Answer
When  multiplying numbers together, exponets are added.

For example, 2²·2³ = 2*2 * 2*2*2 = 4*8 = 32.
This is the same as multiply 5-2's together, and 2+3 = 5.

Similary, 5²·5² = 5*5 * 5*5 = 25*25 = 625.
This is the same as 4-5's all multiplied together, and the exonet is 2+2 = 4.

So, when lookiing at squareroots, we want somenumber times itself that gives the number.
Since all numbers to the 1st are themselves, a squareroot would be two numbers two the 1/2
(or 0.5) that were multiplied together.

It is known that 7*7 = 49, so 49^0.5 = 7.
In this way, 7*7 = (49^0.5)(49^0.5) = 49^(0.5+0.5) = 49^1 = 49.

If finding cuberoots, that is the same as the number to the 1/3 power.
For example, 64^(1/3) = 4, since 4*4*4 = 64.

In this way, any fraction can be used in the exponet.
As another example, take 128^(1/7).
Since 2*2*2*2*2*2*2 = 128 { and that was 7 2's), 128^(1/7) = 2.

As another one, take 15,625^(1/6).
Since 5*5*5*5*5*5 = 15,625 { 6 5's }, 15,625^(1/6) = 5.

The powers of 2 are 2, 4, 8, 16, 32, 64, 128, 256, 512, ...
and in power form they are 2^1, 2^2, 2^3, 2^4, 2^5, 2^6, 2^7, 2^8, andD 2^9, respectively.
This means that the squareroot(4) = 4^0.5 = 2;
cuberoot(8) = 8^(1/3) = 2, since 3-2's multiply out to 8;
fourthroot(16) = 16^0.25 = 2 since 4-2's multiply out to 16.
fifthroot(32) = 32^(1/5) = 32^0.2 = 2, since 5-2's multiply to 32.
..
.
.
ninthroot(512) = 2, since 9-2's multiply to 512.


When dealing with log(x) base 2, that is the power that 2 needs to be taken to.
For example, log(2) = 1, log(4) = 2, log(8) = 3, log(16) = 4, log(32) = 5, etc.

The number e can be found to be the number such that if f(x) = e^x, then f'(x) = e^x.
If f(x) = a^x, then f'(x) = (a^x)ln(a).  When a=e, the ln(e) is 1.
It also arises in interest when dealing with continuous compounding,
biology when computing half-lifes, carbon-dating by dealing with the half-life of carbon,
and many other areas.