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aroub

  • 2 years ago

Simplify using the law of exponents:

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

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

  3. aroub
    • 2 years ago
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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 :)

  4. Jonask
    • 2 years ago
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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

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

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

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

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

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

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

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

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

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

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

  15. Jonask
    • 2 years ago
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    can you illustrate

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

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

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

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

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

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

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

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

  24. aroub
    • 2 years ago
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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 >.<

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

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

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

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

  29. aroub
    • 2 years ago
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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

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

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

  32. aroub
    • 2 years ago
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    Yup!

  33. jhonyy9
    • 2 years ago
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    ok good luck bye

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

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

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

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