additive law: working with negative powers, etc.

crappiefisher26

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Jul 23, 2006
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Additive Law of Exponents in a Multivariate Monomial

x^2y^-2*x^5yz^5*9x^-2y^-4

(9)(x^2*x^5*x^-2)(y^-2*y^-4)(z^5)

9x^2+5-2 y^-2-4 z^5

= 9x^5y^-6z^5

9y^-6z^5/x^5 thats my answer.

heres the correct answer

9x^5z^5/y^5
 

pka

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Jan 29, 2005
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\(\displaystyle \L
\left( {x^2 y^{ - 2} } \right)\left( {x^5 y^1z^5 } \right)\left( {9x^{ - 2} y^{ - 4} } \right) = 9\left( {x^{2 + 5 - 2} } \right)\left( {y^{ - 2 + 1 - 4} } \right)\left( {z^5 } \right) = 9x^5 y^{ - 5} z^5 =\frac{9 x^5 z^5}{y^{5}}\)
 

crappiefisher26

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This is the same type but im not sure how you would set it up.

(x^2/y^3z)*(xy/z^4)*(y^2z^3/x^9)
 

Denis

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Feb 17, 2004
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crappiefisher26 said:
This is the same type but im not sure how you would set it up.
(x^2/y^3z)*(xy/z^4)*(y^2z^3/x^9)
Bring the denominators to the top by changing the sign of the power;
like x^2 / (y^3 z^1) becomes x^2 y^(-3) x^(-1);
then complete as pka showed you.

Your 1st term (x^2/y^3z) requires another set of brackets: (x^2/(y^3z))
As example:
20 / 5 * 2 = 4 * 2 = 8
20 / (5 * 2) = 20 / 10 = 2 : capish :?: :shock:
 

crappiefisher26

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so i would set the problem up like this correct
im confused now.

sorry, is there any way to explain it more thoroughly.

Thanks.
 

pka

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Jan 29, 2005
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7,830
crappiefisher26 said:
so i would set the problem up like this correct
im confused now. sorry, is there any way to explain it more thoroughly.
Man crappiefisher26 you are having far too much trouble with all of this.
It just is not that hard. Do you have a teacher/instructor?
Do you have someone you can go to and have a good sit-down talk to?
You really need to do that before you find yourself totally lost.
If you don’t do that, you may find yourself beyond repair.
 
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