Question

Simplify using the law of exponents:

Answer 1

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

Answer 2

\[c^{m/n}=\sqrt[n]{c^m}\]

Answer 3

- 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 :)

Answer 4

\[\frac{ c^{5/6}d^{1/6} }{ c^{3/4}d^{3/4} }\]
so we convert any radical to exponent using the rule above

Answer 5

The only law that'd help nice would be:\[\large \rm \sqrt[n]{x^m} = x^{m \over n}\]

Answer 6

I know that rule! But then what?

Answer 7

Also, you gotta use:\[\rm {a^x \over a^y} = {a^{x - y}}\]and\[\rm {a^x}\cdot {a^y} = a^{x + y}\]

Answer 8

Then, as JonasK pointed.

Answer 9

\[(\frac{ c^{5/6} }{ c^{3/4} })(\frac{ d^{1/6} }{ d^{3/4} })\]

Answer 10

can you use the rule by @ParthKohli \[\frac{ a^m }{ a^n }=a^{m-n}\]

Answer 11

Jonask, just continue solving please :) All I need is the steps!

Answer 12

\[c^{5/6-3/4}d^{1/6-3/4}\]

Answer 13

same base=subtract powers

Answer 14

cant you just add the numerator with numerator and the denominator with the denominator?

Answer 15

can you illustrate

Answer 16

no we can only addif they are both having same base

Answer 17

cd^5/6+1/6

Answer 18

they have the same base

Answer 19

no c and d are not the same variable \[\frac{ c^n*c^m }{ d^nd^m }=\frac{ c^{m+n} }{ d^{m+n} }\]

Answer 20

Ahh, that's true! Sorry :]

Answer 21

so what is our answer

Answer 22

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

Answer 23

@aroub sorry for this question ,,so why is this like difficile for you ?

Answer 24

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 >.<

Answer 25

,,law of exponents" what method of solving is this ?

Answer 26

You convert EVERYTHING to exponents!

Answer 27

yes than this mean very very easy

Answer 28

begin with sqrtx =x^(1/2)

Answer 29

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

Answer 30

or

a^(3/2)
------- = a^((3/2)-(1/3))
a^(1/3)

right ?

Answer 31

so and i think this will be the law of exponents method sure

yes ?

Answer 32

Yup!

Answer 33

ok

good luck

bye

Answer 34

Thanks! Bye =D

Answer 35

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
\[\sqrt[12]{\frac{C}{D^7}}\]

Answer 36

That's the answer that I actually wanted! @phi thanks a looooottt ^_^