gabriel72alv
(-3w^3 x^-6)^3
Write your answer using only positive exponents.
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RolyPoly
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\[x^{-a} = \frac{1}{x^a}\]\[(xy)^m = x^my^m\]
gabriel72alv
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Please help? I still dont understand? ):
jim_thompson5910
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how far did you get gabriel72alv?
gabriel72alv
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I couldnt figure out how to put it in the equation I needed to solve it? ):
jim_thompson5910
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I'll do a similar example (just with different numbers)
\[\Large \left(-2w^{5}x^{-8}\right)^{4}\]
\[\Large \left(-2\right)^{4}\left(w^{5}\right)^{4}\left(x^{-8}\right)^{4}\]
\[\Large \left((-2)^{1}\right)^{4}\left(w^{5}\right)^{4}\left(x^{-8}\right)^{4}\]
\[\Large (-2)^{1*4}w^{5*4}x^{-8*4}\]
\[\Large (-2)^{4}w^{20}x^{-32}\]
\[\Large (-2)^{4}w^{20}\frac{1}{x^{32}}\]
\[\Large 16w^{20}\frac{1}{x^{32}}\]
\[\Large \frac{16w^{20}}{x^{32}}\]
gabriel72alv
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So would I get (-3w^3/x^6)^3 ?
jim_thompson5910
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Keep going. So far, so good.
gabriel72alv
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Theres still another step? o:
jim_thompson5910
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Yes there are more steps.
gabriel72alv
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How would you work it out to get to the next step though? I understand that.
jim_thompson5910
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You would apply the outermost exponent to each term inside
jim_thompson5910
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In other words, you would cube each piece.
gabriel72alv
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So basically would it be, (-3w^3/x^6) times (-3w^3/x^6) times (-3w^3/x^6) ?
jim_thompson5910
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Yes pretty much
jim_thompson5910
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So that explains why, for instance, in the denominator you will have
x^6*x^6*x^6 = x^(6+6+6) = x^(18)
jim_thompson5910
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A shortcut is to use this rule
\[\Large (x^m)^n = x^{m*n}\]
gabriel72alv
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So would I get for my answer to that is -27w^9/x^18 ?
jim_thompson5910
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You are 100% correct. Nice work.
gabriel72alv
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Nicee! Haa, So thats my answer? xD
jim_thompson5910
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Yes it is.
gabriel72alv
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Awesome, Thank you so much jim_thompson5910 ! (:
jim_thompson5910
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you're welcome