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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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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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