aroub
  • aroub
Simplify using the law of exponents:
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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aroub
  • aroub
\[\frac{ \sqrt[6]{c^5d} }{ \sqrt[4]{c^3d^3} }\]
anonymous
  • anonymous
\[c^{m/n}=\sqrt[n]{c^m}\]
aroub
  • aroub
- 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 :)

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More answers

anonymous
  • anonymous
\[\frac{ c^{5/6}d^{1/6} }{ c^{3/4}d^{3/4} }\] so we convert any radical to exponent using the rule above
ParthKohli
  • ParthKohli
The only law that'd help nice would be:\[\large \rm \sqrt[n]{x^m} = x^{m \over n}\]
aroub
  • aroub
I know that rule! But then what?
ParthKohli
  • ParthKohli
Also, you gotta use:\[\rm {a^x \over a^y} = {a^{x - y}}\]and\[\rm {a^x}\cdot {a^y} = a^{x + y}\]
ParthKohli
  • ParthKohli
Then, as JonasK pointed.
anonymous
  • anonymous
\[(\frac{ c^{5/6} }{ c^{3/4} })(\frac{ d^{1/6} }{ d^{3/4} })\]
anonymous
  • anonymous
can you use the rule by @ParthKohli \[\frac{ a^m }{ a^n }=a^{m-n}\]
aroub
  • aroub
Jonask, just continue solving please :) All I need is the steps!
anonymous
  • anonymous
\[c^{5/6-3/4}d^{1/6-3/4}\]
anonymous
  • anonymous
same base=subtract powers
aroub
  • aroub
cant you just add the numerator with numerator and the denominator with the denominator?
anonymous
  • anonymous
can you illustrate
anonymous
  • anonymous
no we can only addif they are both having same base
aroub
  • aroub
cd^5/6+1/6
aroub
  • aroub
they have the same base
anonymous
  • anonymous
no c and d are not the same variable \[\frac{ c^n*c^m }{ d^nd^m }=\frac{ c^{m+n} }{ d^{m+n} }\]
aroub
  • aroub
Ahh, that's true! Sorry :]
anonymous
  • anonymous
so what is our answer
anonymous
  • anonymous
\[c^{-2/24}d^{-14/12}\] is the answer lets try this one\[\frac{ x^3y^{-2} }{ \sqrt[3]{x^2y^2}}\]
jhonyy9
  • jhonyy9
@aroub sorry for this question ,,so why is this like difficile for you ?
aroub
  • aroub
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 >.<
jhonyy9
  • jhonyy9
,,law of exponents" what method of solving is this ?
aroub
  • aroub
You convert EVERYTHING to exponents!
jhonyy9
  • jhonyy9
yes than this mean very very easy
jhonyy9
  • jhonyy9
begin with sqrtx =x^(1/2)
aroub
  • aroub
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
jhonyy9
  • jhonyy9
or a^(3/2) ------- = a^((3/2)-(1/3)) a^(1/3) right ?
jhonyy9
  • jhonyy9
so and i think this will be the law of exponents method sure yes ?
aroub
  • aroub
Yup!
jhonyy9
  • jhonyy9
ok good luck bye
aroub
  • aroub
Thanks! Bye =D
phi
  • phi
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 \[\sqrt[12]{\frac{C}{D^7}}\]
aroub
  • aroub
That's the answer that I actually wanted! @phi thanks a looooottt ^_^

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