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Simplify using the law of exponents:

Mathematics
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\[\frac{ \sqrt[6]{c^5d} }{ \sqrt[4]{c^3d^3} }\]
\[c^{m/n}=\sqrt[n]{c^m}\]
- 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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Other answers:

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

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