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Old 09-04-08, 11:37 AM   #1
UnsafeAlpine
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What the crap (more algebra)

(8x^-12 y^24/27z^-18)^1/3

Simplify each expression leaving only positive exponents.

I don't even know where to start

can I multiply each exponent by 1/3?
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Old 09-04-08, 11:38 AM   #2
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Don't let jschen see this. He'll pull out the math book on you.
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Old 09-04-08, 11:39 AM   #3
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I'm cool with that...This is just the stuff I'm already supposed to know Haven't had a math class in 14 years and it wasn't this advanced...
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Old 09-04-08, 11:40 AM   #4
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start out by cubing the expression to get rid of the exponent.
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Old 09-04-08, 11:46 AM   #5
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so it would be like this?

(8x^-12y^24/27z^-28)

I don't see how you can do that, though.
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Old 09-04-08, 11:53 AM   #6
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you are trying to establish, what is the cube root of that equation. Calculate the cube root of the top and the bottom separately.
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Old 09-04-08, 11:56 AM   #7
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pssst! The answers are in the back of the book.
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Old 09-04-08, 11:56 AM   #8
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Just distribute the 1/3 exponent through, and remember what a negative exponent means (and how you make it "positive")
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Old 09-04-08, 11:57 AM   #9
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*flicks Taerom in forehead* Shhhhh!!
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Old 09-04-08, 11:57 AM   #10
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Ok. so the top

(8x^-12 y^24)^1/3

(8x^-4 y^8) ?
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Old 09-04-08, 11:58 AM   #11
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I wish I had a book. this is a worksheet...blargh
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Old 09-04-08, 12:00 PM   #12
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Since (x/y)^ 1/3 = (X ^ 1/3) / (y ^ 1/3)

8x^-12 y^24/27z^-18)^1/3 = [(8x^-12 y^24)^1/3] / [(27z^-18)^1/3]

It should be pretty straight forward to simplify the numerator and denominator...

Then figure out how to make the negative exponents positive...
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Old 09-04-08, 12:01 PM   #13
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Ok. so the top

(8x^-12 y^24)^1/3

(8x^-4 y^8) ?
All except you need to do something about that first 8...
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Old 09-04-08, 12:02 PM   #14
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cube root of 8 =2
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Old 09-04-08, 12:03 PM   #15
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Then figure out how to make the negative exponents positive...
zoloft, prozac?
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Old 09-04-08, 12:05 PM   #16
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zoloft, prozac?
No, that way they just don't care that they're negative...
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Old 09-04-08, 12:06 PM   #17
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I'm possibly getting a job with a local university in which case i get to take classes for free. I think math is my next area of study
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Old 09-04-08, 12:07 PM   #18
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Huh?
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Please dont outsmart the censor. That is a very expensive censor and every time one of you guys outsmart it it makes someone at the home office feel bad. We dont wanna do that. So dont cleverly disguise bad words.
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Old 09-04-08, 12:08 PM   #19
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oh yeah, right. all the forms get raised to the power of 1/3. cool so the top is

(2x^-3 y^8)/(3z^-6)

So now I need to figure out how to flip the exponent signs. Can I do this?

(2z^6 y^8)/(3x^3) ?
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Old 09-04-08, 12:10 PM   #20
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Indeed you can...

Since x^-1 = 1/x
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Old 09-04-08, 12:13 PM   #21
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WooHoo! that's 2 out of...let me see here...er...20...
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Old 09-04-08, 12:22 PM   #22
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What makes these easier is finding the set of (mathematical) tools that you can use for each situation

So for exponents, a few tools include:

for any value of a and b

a^b + a^b = 2*(a^b)

a^b * a^1 = a^(b+1) which is the same as a^b * a^b = a^2b

a^-b = 1/(a^b)

a^(1/2) = the square root of a

a^(1/3) = the cude root of a

a^(2/3) = the cube root of a^2

and so on
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Old 09-04-08, 12:23 PM   #23
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Sorry that there are so many formulas, but i find that if you know how to use the formulas, you can dissect the problem into smaller, easier problems
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Old 09-04-08, 12:29 PM   #24
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Yeah, Cool thanks...I'll have to write them down (it's easier to see it that way)
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Old 09-04-08, 12:38 PM   #25
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You seem like such a nice kid from your posts. Why would you want to do this to yourself?
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