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aroub

Simplify using the law of exponents:

  • one year ago
  • one year ago

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

    • one year ago
  2. Jonask
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    \[c^{m/n}=\sqrt[n]{c^m}\]

    • one year ago
  3. aroub
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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 :)

    • one year ago
  4. Jonask
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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

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

    • one year ago
  6. aroub
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    I know that rule! But then what?

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

    • one year ago
  8. ParthKohli
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    Then, as JonasK pointed.

    • one year ago
  9. Jonask
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    \[(\frac{ c^{5/6} }{ c^{3/4} })(\frac{ d^{1/6} }{ d^{3/4} })\]

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

    • one year ago
  11. aroub
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    Jonask, just continue solving please :) All I need is the steps!

    • one year ago
  12. Jonask
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    \[c^{5/6-3/4}d^{1/6-3/4}\]

    • one year ago
  13. Jonask
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    same base=subtract powers

    • one year ago
  14. aroub
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    cant you just add the numerator with numerator and the denominator with the denominator?

    • one year ago
  15. Jonask
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    can you illustrate

    • one year ago
  16. Jonask
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    no we can only addif they are both having same base

    • one year ago
  17. aroub
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    cd^5/6+1/6

    • one year ago
  18. aroub
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    they have the same base

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

    • one year ago
  20. aroub
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    Ahh, that's true! Sorry :]

    • one year ago
  21. Jonask
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    so what is our answer

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

    • one year ago
  23. jhonyy9
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    @aroub sorry for this question ,,so why is this like difficile for you ?

    • one year ago
  24. aroub
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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 >.<

    • one year ago
  25. jhonyy9
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    ,,law of exponents" what method of solving is this ?

    • one year ago
  26. aroub
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    You convert EVERYTHING to exponents!

    • one year ago
  27. jhonyy9
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    yes than this mean very very easy

    • one year ago
  28. jhonyy9
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    begin with sqrtx =x^(1/2)

    • one year ago
  29. aroub
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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

    • one year ago
  30. jhonyy9
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    or a^(3/2) ------- = a^((3/2)-(1/3)) a^(1/3) right ?

    • one year ago
  31. jhonyy9
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    so and i think this will be the law of exponents method sure yes ?

    • one year ago
  32. aroub
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    Yup!

    • one year ago
  33. jhonyy9
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    ok good luck bye

    • one year ago
  34. aroub
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    Thanks! Bye =D

    • one year ago
  35. phi
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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 \[\sqrt[12]{\frac{C}{D^7}}\]

    • one year ago
  36. aroub
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    That's the answer that I actually wanted! @phi thanks a looooottt ^_^

    • one year ago
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