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## JessicaNicoleCraven k^3(k7/5)^-5 one year ago one year ago Edit Question Delete Cancel Submit

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1. angela210793
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|dw:1352048933|dw:1352048968202:dw|095:dw|

• one year ago
2. CliffSedge
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? Is $$\large k^3 (\frac{k^7}{5})^{-5}$$ ?

• one year ago
3. angela210793
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:O where did my solution go :O

• one year ago
4. CliffSedge
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^ ? I dunno, looks like you were trying to post a drawing.

• one year ago
5. JessicaNicoleCraven
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I am so confused. Its suppose to be k with an exponent of 3, times k(7/5) with an exponent of negative 5.

• one year ago
6. CliffSedge
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Then is it $$\large k^3 \cdot (k^{7/5})^{-5}$$ ?

• one year ago
7. JessicaNicoleCraven
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Yeah.

• one year ago
8. CliffSedge
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Ok. Order of operations, handle the exponent first. A power-to-a-power, multiply the exponents.

• one year ago
9. angela210793
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All u have to know is that x^a*x^b=x^(a+b) (x^a)^b=x^ab and x^-a=1/x^a so u'll have |dw:1352049321095:dw|

• one year ago
10. CliffSedge
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$$\huge \rightarrow k^3 \cdot k^{(\frac{7}{5} \cdot -5)}$$ $$\huge \rightarrow k^3 \cdot k^{-7}$$

• one year ago
11. JessicaNicoleCraven
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Oh wow, thats a lot more simple than I was trying to make it. Thank you!

• one year ago
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