what is the concept around an exponent or root that is a decimal.

its cool for example 125^2 is telling me "times me by myself"

125^3 is telling me "times me by myself and do it again!"

but what is 125^2.5 telling me to do?

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- Nov 7th 2007, 05:55 PMflacck42decimal exponents and roots, whats the big idea?
what is the concept around an exponent or root that is a decimal.

its cool for example 125^2 is telling me "times me by myself"

125^3 is telling me "times me by myself and do it again!"

but what is 125^2.5 telling me to do? - Nov 7th 2007, 06:06 PMtopsquark
Some of this is easy, some of it isn't.

Decimals that can be readily turned into fractions have an immediate interpretation. For example, you can show that

$\displaystyle a^{1/n} = \sqrt[n]{a}$

So your example

$\displaystyle 125^{2.5} = 125^{5/2} = \sqrt{125^5}$

If the decimal is something like $\displaystyle \sqrt{2}$ (ie. irrational) I am much more hazy about the interpretation. We can, however, do this:

$\displaystyle a^{\sqrt{2}} = a^{1.41421...}$

$\displaystyle = a^{1 + 0.4 + 0.01 + 0.004 + 0.0002 + 0.00001 + ...}$

$\displaystyle = (a^1)(a^{4/10})(a^{1/100})(a^{4/1000})(a^{2/10000})(a^{1/100000})...$

But that's still pretty messy to work with.

-Dan - Nov 7th 2007, 06:06 PMJhevon
fractional exponents get a little weird.

Note that $\displaystyle 125^{2.5} = 125^{5/2}$

now what this says is "times me by myself and do it again and again and again, and while you at it, take my square root (but that's if you interpret me as $\displaystyle \left( 125^5\right)^{1/2}$). If you interpret me as $\displaystyle \left( 125^{1/2} \right)^5$, then take my square root and then raise me to the 5th power"

thus, $\displaystyle 125^{2.5} = 125^{5/2} = \sqrt{125^5} = \left( \sqrt{125} \right)^5$

if the power is irrational (cannot be expressed as a fraction, then, you need your calculator)

see post #3 here for more info

EDIT: Thanks for letting me waste my time answering this, Dan :p - Nov 7th 2007, 06:17 PMtopsquark
- Nov 7th 2007, 06:23 PMJhevon
hehe, you know you caused him to pass you:D

i will try, this thing stops me from adding rep sometimes. i've been trying to add rep to Krizalid forever (i just tried to add rep to you and it wouldn't allow me to)

TPH went offline, hopefully next time he logs on he remembers how many points he had before i added rep - Nov 7th 2007, 06:26 PMtopsquark
- Nov 7th 2007, 07:30 PMflacck42
question two

should i rack my brain mulling over your concepts or would they come to me eventually (through osmosis?) and chillax a bit?

would i be in better position to sink these concepts after i learned calculus? - Nov 7th 2007, 07:42 PMJhevon
definitely rack your brain. the osmosis takes care of itself. you'd find that sometimes after racking your brain for a long time, you will give up and go to something else, when suddenly, BAM! with no warning, it comes to you. (your subconscious will get the message that figuring this out is important to you because you were racking your brain, and so it works on it while you're not noticing)

Quote:

would i be in better position to sink these concepts after i learned calculus?

- Nov 7th 2007, 08:47 PMflacck42
got it:

125^2.5 says

split 125 into two equal parts that when multiplied equals 125:

$\displaystyle 11.2*11.2=125$

take one of those parts (11.2) and multiply that by 125 and 125 again.

likewise 125^2.75 says

split 125 into four parts that when multiplied together equals 125

$\displaystyle 3.3*3.3*3.3*3.3=125$

take three of those parts $\displaystyle 3.3*3.3*3.3=37.4$ and multiply by 125 and 125 again.

im going to watch cartoons. - Nov 7th 2007, 09:33 PMangel.white
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Or, if you plug your numbers into the equation, and they come out to roughly 50 points, then it is probably accurate. - Jan 15th 2008, 04:31 AMqspeechc
What about 5^(sqrt(2))?

- Jan 15th 2008, 06:47 AMtopsquark
Do you mean how to calculate it?

Probably the simplest thing to do is to get the decimal representation of $\displaystyle \sqrt{2} = 1.41421...$.

Thus

$\displaystyle 5^{\sqrt{2}} = 5^{1.41421...} = (5^1)(5^{0.4})(5^{0.01})(5^{0.004})(5^{0.0002})(5^ {0.00001})...$

$\displaystyle = 5 \cdot 5^{4/10} \cdot 5^{1/100} \cdot 5^{4/1000} \cdot 5^{2/10000} \cdot 5^{1/100000} \cdot ~...$

So you can calculate the first several terms in the product and get at least a decent approximation to it. (Note as the series progresses to smaller and smaller fractions $\displaystyle 5^n$ approaches 1.) But you can't get an exact value for it.

Plugging it into my calculator gives $\displaystyle 5^{\sqrt{2}} \approx 9.73852$

-Dan