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chrisGA

  • one year ago

Evaluate fourth root of 9 multiplied by square root of 9 over the fourth root of 9 to the power of 5.

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  1. chrisGA
    • one year ago
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    9 to the power of negative 1 over 2 9 to the power of negative 1 over 4 9 92

  2. anonymous
    • one year ago
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    If I understand correctly, the question is to simplify\[\frac{ 9^{\frac{ 1 }{ 4 }} \times 9^{\frac{ 1 }{ 2 }} }{ \left( 9^{\frac{ 1 }{ 4 }} \right)^{5} }\]Is this correct?

  3. chrisGA
    • one year ago
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  4. anonymous
    • one year ago
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    OK. When working with exponents, many times it is easier to express them as fractions rather than radicals. My statement is equivalent to yours. Do you understand the fractional exponents?

  5. chrisGA
    • one year ago
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    yes a lil

  6. anonymous
    • one year ago
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    For example,\[\sqrt{x}=x ^{\frac{ 1 }{ 2 }}\]\[\sqrt[3]{x}=x ^{\frac{ 1 }{ 3 }}\]\[\sqrt[4]{x}=x ^{\frac{ 1 }{ 4 }}\]\[\sqrt[4]{x ^{5}}=\left( x ^{5} \right)^{\frac{ 1 }{ 4 }}=x ^{\frac{ 5 }{ 4 }}\]

  7. anonymous
    • one year ago
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    So, with the expression written with fractional exponents, simplify the numerator first. The rule when multiplying powers with the same base is to ADD the exponents. Then, divide the numerator by the denominator, The rule when dividing powers with the same base is that you SUBTRACT the exponents. This help?

  8. chrisGA
    • one year ago
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    it does a lil bit what would the answer be?

  9. anonymous
    • one year ago
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    For example,\[\frac{ 3^{\frac{ 1 }{ 4 }}\times 3^{\frac{ 1 }{ 2 }} }{ 3^{\frac{ 7 }{ 4 }} }=\frac{ 3^{\frac{ 3 }{ 4 }} }{ 3^{\frac{ 7 }{ 4 }} }=3^\left( \frac{ 3 }{ 4 } -\frac{ 7 }{ 4 }\right)=3^{-\frac{ 4 }{ 4 }}=3^{-1}\]

  10. anonymous
    • one year ago
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    Sorry for taking so long with the typing. I can't give you the answer, but I can check yours. What do you get?

  11. chrisGA
    • one year ago
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    it ok and i got B? is that correct

  12. anonymous
    • one year ago
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    Don't think so. Let's take it one step at a time. What do you get when you simplify just the numerator?

  13. chrisGA
    • one year ago
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    9

  14. anonymous
    • one year ago
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    \[9^{\frac{ 1 }{ 4 }}\times 9^{\frac{ 1 }{ 2 }} = 9^{\left( \frac{ 1 }{ 4 } +\frac{ 1 }{ 2 }\right)}=?\]

  15. chrisGA
    • one year ago
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    is it C?

  16. anonymous
    • one year ago
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    Don't think so. How about simplifying the numerator first, as above. What would you get?

  17. anonymous
    • one year ago
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    What is 1/4 + 1/2 ?

  18. chrisGA
    • one year ago
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    3/4

  19. anonymous
    • one year ago
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    Hello? Do you still want my help?

  20. chrisGA
    • one year ago
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    3/4

  21. anonymous
    • one year ago
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    Right on! So the numerator simplifies to\[9^{\frac{ 3 }{ 4 }}\]Now simplify the denominator. It is a power raise to another exponent. The rule in this case is to MULTIPLY the exponents. So\[\left(9^{\frac{ 1 }{ 4 }} \right)^{5}=9^{?}\] What do you get?

  22. chrisGA
    • one year ago
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    2?

  23. anonymous
    • one year ago
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    No. Keep the base (9) and multiply just the exponents together.\[\left( 9^{\frac{ 1 }{ 4 }} \right)^{5} = 9^{\left( \frac{ 1 }{ 4 } \times5\right)} = 9^{?}\]

  24. anonymous
    • one year ago
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    In other words, what is 1/4 x 5 ?

  25. chrisGA
    • one year ago
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    5/4?

  26. anonymous
    • one year ago
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    Perfect. Now your problem has been simplified to \[\frac{ 9^{\frac{ 3 }{ 4 }} }{ 9^{\frac{ 5 }{ 4 }} }\]When dividing powers with the same base, the rule is to keep the base and SUBTRACT the exponents. Therefore\[\frac{ 9^{\frac{ 3 }{ 4 }} }{ 9^{\frac{ 5 }{ 4 }} } = 9^{\left( \frac{ 3 }{ 4 } -\frac{ 5 }{ 4 }\right)} = 9^{?}\]

  27. anonymous
    • one year ago
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    In other words, what is 3/4 - 5/4 ?

  28. chrisGA
    • one year ago
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    -1/2

  29. anonymous
    • one year ago
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    Terrific. That's your answer.\[9^{-\frac{ 1 }{ 2 }}\] Lot of small step. And lots of rules for working with exponents. Well done!

  30. chrisGA
    • one year ago
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    thank you very much

  31. anonymous
    • one year ago
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    You're welcome. A bit of practice, and you'll be a master.

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