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anonymous

  • 5 years ago

243^2/5

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  1. anonymous
    • 5 years ago
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    9 as it is (243^1/5)^2

  2. anonymous
    • 5 years ago
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    yeah i figured it out

  3. anonymous
    • 5 years ago
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    thanks

  4. anonymous
    • 5 years ago
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    \[243^{2/5}=\sqrt[5]{243^2}\]\[\sqrt[5]{243^2}=\sqrt[5]{59049}\]\[\sqrt[5]{59049}=9\]

  5. anonymous
    • 5 years ago
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    |dw:1326864018846:dw|

  6. anonymous
    • 5 years ago
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    \[\sqrt[4]{x ^{16}y^4}=x^4y\]Take out 1 variable for every factor of 4.

  7. anonymous
    • 5 years ago
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    oh ok thanks

  8. anonymous
    • 5 years ago
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    |dw:1326864378417:dw|

  9. anonymous
    • 5 years ago
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    what about this type

  10. anonymous
    • 5 years ago
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    Is that\[(x ^{1/2}y^3)^2\sqrt{x^2}\]?

  11. anonymous
    • 5 years ago
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    yes that is. sorry bad drawing

  12. anonymous
    • 5 years ago
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    It's cool.

  13. anonymous
    • 5 years ago
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    We can split that up into two parts\[(x ^{1/2}y^3)^2\]And\[\sqrt{x^2}\]We'll do the first part, there's a rule\[(x^a)^b=x ^{a \times b}\]So...\[(x ^{1/2}y^3)^2=xy^6\]Also, when you have a square being square rooted they cancel out.\[\sqrt{x^2}=x\]Finally\[xy^6x=x^2y^6\]

  14. anonymous
    • 5 years ago
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    ok thanks

  15. anonymous
    • 5 years ago
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    |dw:1326865345268:dw| so fill in the blank

  16. anonymous
    • 5 years ago
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    \[256^{a/4}=4\]\[\sqrt[4]{256^a}=4\]\[(\sqrt[4]{256^a})^4=(4)^4\]\[256^a=256\]\[a=1\]

  17. anonymous
    • 5 years ago
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    |dw:1326865703133:dw| so for this what would be the steps

  18. anonymous
    • 5 years ago
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    \[243{}^{\wedge}(2/5) = 9 \]\[243{}^{\wedge}2/5 = \frac{59049}{5} \]

  19. anonymous
    • 5 years ago
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    \[225^{1/a}=15\]\[\sqrt[a]{225^1}=15\]Any number/variable to the 1 power is that number/variable.\[x^1=x\]\[387461873654^1=387461873654\]\[\sqrt[a]{225}=15\]\[(\sqrt[a]{225})^a=(15)^a\]\[225=15^a\]\[15 \times 15=225\]Therefore\[a=2\]

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