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aroub Group Title

Simplify using the law of exponents:

  • 2 years ago
  • 2 years ago

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  1. aroub Group Title
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    \[\frac{ \sqrt[6]{c^5d} }{ \sqrt[4]{c^3d^3} }\]

    • 2 years ago
  2. Jonask Group Title
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    \[c^{m/n}=\sqrt[n]{c^m}\]

    • 2 years ago
  3. aroub Group Title
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    - I never actually solved a question using " the laws of exponents" I just convert the final answer.. So, if you could just show me how or the steps of solving it please :)

    • 2 years ago
  4. Jonask Group Title
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    \[\frac{ c^{5/6}d^{1/6} }{ c^{3/4}d^{3/4} }\] so we convert any radical to exponent using the rule above

    • 2 years ago
  5. ParthKohli Group Title
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    The only law that'd help nice would be:\[\large \rm \sqrt[n]{x^m} = x^{m \over n}\]

    • 2 years ago
  6. aroub Group Title
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    I know that rule! But then what?

    • 2 years ago
  7. ParthKohli Group Title
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    Also, you gotta use:\[\rm {a^x \over a^y} = {a^{x - y}}\]and\[\rm {a^x}\cdot {a^y} = a^{x + y}\]

    • 2 years ago
  8. ParthKohli Group Title
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    Then, as JonasK pointed.

    • 2 years ago
  9. Jonask Group Title
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    \[(\frac{ c^{5/6} }{ c^{3/4} })(\frac{ d^{1/6} }{ d^{3/4} })\]

    • 2 years ago
  10. Jonask Group Title
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    can you use the rule by @ParthKohli \[\frac{ a^m }{ a^n }=a^{m-n}\]

    • 2 years ago
  11. aroub Group Title
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    Jonask, just continue solving please :) All I need is the steps!

    • 2 years ago
  12. Jonask Group Title
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    \[c^{5/6-3/4}d^{1/6-3/4}\]

    • 2 years ago
  13. Jonask Group Title
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    same base=subtract powers

    • 2 years ago
  14. aroub Group Title
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    cant you just add the numerator with numerator and the denominator with the denominator?

    • 2 years ago
  15. Jonask Group Title
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    can you illustrate

    • 2 years ago
  16. Jonask Group Title
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    no we can only addif they are both having same base

    • 2 years ago
  17. aroub Group Title
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    cd^5/6+1/6

    • 2 years ago
  18. aroub Group Title
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    they have the same base

    • 2 years ago
  19. Jonask Group Title
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    no c and d are not the same variable \[\frac{ c^n*c^m }{ d^nd^m }=\frac{ c^{m+n} }{ d^{m+n} }\]

    • 2 years ago
  20. aroub Group Title
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    Ahh, that's true! Sorry :]

    • 2 years ago
  21. Jonask Group Title
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    so what is our answer

    • 2 years ago
  22. Jonask Group Title
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    \[c^{-2/24}d^{-14/12}\] is the answer lets try this one\[\frac{ x^3y^{-2} }{ \sqrt[3]{x^2y^2}}\]

    • 2 years ago
  23. jhonyy9 Group Title
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    @aroub sorry for this question ,,so why is this like difficile for you ?

    • 2 years ago
  24. aroub Group Title
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    No, it's okay! I'll tell you why, right there's two ways of solving this? First you can either solve it by keeping the radicals and then the final answer you can convert it OR by using the laws of exponents.. I always solve it using the first one -I find it easier. But then I found it that I have to solve it using the laws of exponents, so here I am >.<

    • 2 years ago
  25. jhonyy9 Group Title
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    ,,law of exponents" what method of solving is this ?

    • 2 years ago
  26. aroub Group Title
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    You convert EVERYTHING to exponents!

    • 2 years ago
  27. jhonyy9 Group Title
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    yes than this mean very very easy

    • 2 years ago
  28. jhonyy9 Group Title
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    begin with sqrtx =x^(1/2)

    • 2 years ago
  29. aroub Group Title
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    Lol, no to me -.- Well, I know how to convert them to exponents but then I have nooo ideaa! Anyway, thanks everyoonee! I gtg :D

    • 2 years ago
  30. jhonyy9 Group Title
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    or a^(3/2) ------- = a^((3/2)-(1/3)) a^(1/3) right ?

    • 2 years ago
  31. jhonyy9 Group Title
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    so and i think this will be the law of exponents method sure yes ?

    • 2 years ago
  32. aroub Group Title
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    Yup!

    • 2 years ago
  33. jhonyy9 Group Title
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    ok good luck bye

    • 2 years ago
  34. aroub Group Title
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    Thanks! Bye =D

    • 2 years ago
  35. phi Group Title
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    I assume you have the answer, but to finish this post: \[ C^{(\frac{5}{6}-\frac{3}{4})}\cdot D^{(\frac{1}{6}-\frac{3}{4})} \] C's exponent is \[ \frac{5}{6}-\frac{3}{4} =\frac{5}{6} \cdot \frac{2}{2}-\frac{3}{4} \cdot \frac{3}{3}= \frac{10}{12}- \frac{9}{12}= \frac{1}{12}\] D's exponent is \[ \frac{1}{6}-\frac{3}{4} =\frac{1}{6} \cdot \frac{2}{2}-\frac{3}{4} \cdot \frac{3}{3}= \frac{2}{12}- \frac{9}{12}= -\frac{7}{12}\] you now hve \[ C^{\frac{1}{12}}D^{-\frac{7}{12}}= (\frac{C}{D^7})^\frac{1}{12}\] or \[\sqrt[12]{\frac{C}{D^7}}\]

    • 2 years ago
  36. aroub Group Title
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    That's the answer that I actually wanted! @phi thanks a looooottt ^_^

    • 2 years ago
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