Question

Simplify the given expression to rational exponent form, justify each step by identifying the properties of rational exponents used. All work must be shown.

Answer 1

\[\frac{ 1 }{ \sqrt[3]{x} -6}\]

Answer 2

So far I think that I should work on the bottom half of the problem which is \[\sqrt[3]{x}-6\]

Answer 3

which is \[x \frac{ -6 }{ 3 }\]

Answer 4

to make it positive I \[\frac{ 1 }{ x \frac{ 6 }{ 3 } }\]

Answer 5

so I think the answer is \[1/\frac{ 1 }{ x \frac{ 6 }{ 3 } }\]

Answer 6

am I correct?

Answer 7

@Hero can you help me please?

Answer 8

@jim_thompson5910 ?

Answer 9

@Directrix please?

Answer 10

@ganeshie8 please?

Answer 11

so they want you to rationalize the denominator?

Answer 12

they want it to be a rational exponent form? i'm not exactly sure what that means?

Answer 13

well why not use the idea that

\[\large \sqrt[n]{x^m} = x^{m/n}\]

so that means

\[\large \sqrt[3]{x} = x^{1/3}\]

-------------------------------------------------------

so

\[\large \frac{ 1 }{ \sqrt[3]{x} -6}\]

turns into

\[\large \frac{ 1 }{ x^{1/3} -6}\]

but I'm not sure if this is what they're looking for

Answer 14

ah the -6 is an exponent inside the square root

Answer 15

what do you mean?

Answer 16

so that's why I thought it was 1/1/x6/3

Answer 17

so the original expression isn't

\[\large \frac{ 1 }{ \sqrt[3]{x} -6}\]

???

Answer 18

also wouldn't x^6/3 be 2x if simplified?

Answer 19

\[\large x^{6/3} = x^{2/1} = x^2\]

\[\large x^{6/3} = x^2\]

Answer 20

x^2 is NOT the same as 2x

Answer 21

no, I will take a picture of it and ohh you're right

Answer 22

so would it be then 1/1/x^2?

Answer 23

ooh i see now

Answer 24

wahhhhhh

Answer 25

wow

Answer 26

\[\large \frac{ 1 }{ \sqrt[3]{x^{-6}}}\]


\[\large \frac{ 1 }{ \left(x^{-6}\right)^{1/3}}\]


\[\large \frac{ 1 }{ x^{-6*1/3}}\]


\[\large \frac{ 1 }{ x^{-6/3}}\]


\[\large \frac{ 1 }{ x^{-2}}\]


\[\large \frac{ 1 }{ \frac{1}{x^{-2}}}\]


\[\large x^{2}\]


-------------------------------------------------------


So


\[\large \frac{ 1 }{ \sqrt[3]{x^{-6}}}\]


simplifies to


\[\large x^{2}\]

Answer 27

so you can see you are very close

Answer 28

but where did the 1/3 come from?

Answer 29

did you separate them on purpose ?

Answer 30

because

\[\large \sqrt[3]{x} = x^{1/3}\]

we know that

\[\large \sqrt[3]{x^{-6}} = \left(x^{-6}\right)^{1/3}\]

Answer 31

so instead of x^-6/3 you made it that way ahhhh but I get it now wow thank you very much

Answer 32

do you think you can help me with one more problem?

Answer 33

yeah there are multiple ways to get to the answer (there's probably a much easier way, but it depends on the person)

and sure I can help

Answer 34

thank you and here is the problem:

Answer 35

One of your friends sends you a WebMail message asking you to explain how all of the following expressions have the same answer.
\[\sqrt[3]{x^3}\]
\[x \frac{ 1 }{ 3 } * x \frac{ 1 }{ 3 } * x \frac{ 1 }{ 3 }\]
\[\frac{ 1 }{ x^1 }\] ^ the exponent 1 is negative
\[\sqrt[11]{x^5 * x^4 * x^2}\]
Compose a WebMail message back assisting your friend and highlight the names of the properties of exponents when you use them.

Answer 36

so far I understand how the first two have the same answer but I don't understand the last two

Answer 37

when you have a negative exponent, you flip the fraction that it applies to make the exponent positive

so in general

\[\large \left(\frac{a}{b}\right)^{-k} = \left(\frac{b}{a}\right)^{k}\]

notice how a/b flipped to b/a and the exponent changed in sign

Answer 38

so using this rule we can say

\[\large \left(\frac{1}{x^1}\right)^{-1}\]

turns into

\[\large \left(\frac{x^1}{1}\right)^{1}\]

Answer 39

so it would be x^1/1 whicish x^1 which is x but how is that the same as the first two

Answer 40

well the cube root of x^3 is x

this is because the cube root operation cancels out the cubing operation (kinda like how square roots undo squaring or how division undoes multiplication)

Answer 41

example

take 2 and cube it to get 2^3 = 8

then take the cube root of 8 to get 2 back again

----------------

in general, cube any number x to get x^3

then take the cube root of x^3 to get x back again

Answer 42

so that's why

\[\large \sqrt[3]{x^3} = x\]

Answer 43

ahhhh your're right I looked at something wrong so the first three are the and the last one is the same because there is 11 inside the square root and outside

Answer 44

exactly

Answer 45

and the second one is just a different form of the first one

this is because

\[\large \sqrt[3]{x^3} = \left(\sqrt[3]{x}\right)^3\]

Answer 46

ahh I get it all now :D and about the first question I asked you, what properties did you use to solve the problem?

Answer 47

im' not very familiar with the properties yet

Answer 48

which steps exactly are you referring to?

Answer 49

the first and second ones

Answer 50

?

Answer 51

hmm I'm blanking on the proper names and only can only find pages that only list the properties themselves but they don't have names

Answer 52

for instance the rule

\[\large \sqrt[n]{x^m} = x^{m/n}\]

doesn't have an official name I don't think

Answer 53

I could try giving the rule I used for each step

does that work?

Answer 54

yes that would help :)

Answer 55

also if you could also give me the property name of so using this rule we can say

(1 x 1 ) −1


turns into

(x 1 1 ) 1

Answer 56

ok one sec

Answer 57

thank you very much :)

Answer 58

\[\large \frac{ 1 }{ \sqrt[3]{x^{-6}}}\]


\[\large \frac{ 1 }{ \left(x^{-6}\right)^{1/3}}\]


\[\large \frac{ 1 }{ x^{-6*1/3}}\]


\[\large \frac{ 1 }{ x^{-6/3}}\]


\[\large \frac{ 1 }{ x^{-2}}\]


\[\large \frac{ 1 }{ \frac{1}{x^{2}}}\]


\[\large x^{2}\]



Steps done


Step 1: Given expression


Step 2: used rule \(\large \sqrt[n]{x^m} = x^{m/n}\)


Step 3: used rule \(\large \left(x^{a}\right)^{b} = x^{a*b}\)

Note: this rule is may be called "The product of powers property"


Step 4: Multiplication


Step 5: Division


Step 6: Used rule \(\large x^{-k} = \frac{1}{x^k}\)


Step 7: Used rule \(\large \frac{1}{a/b} = \frac{b}{a}\)

Answer 59

THANK YOU VERY MUCH! this helps a lot!

Answer 60

I'm glad it does