Question

Hello guys I need help with this equation
Simplify: 8 -2/3(the -2/3 is an exponent)

Answer 1

LOL sure :)
First deal with the 'negativeness' of the exponent, that brings it to the denominator (with a positive exponent)
\[\Large a^{-m}= \frac1{a^m}\]

Answer 2

\[\huge \frac{1}{8^{\frac23}}\]
Catch me so far?

Answer 3

yes I got that part but I flipped it so the 1 would be at the bottom after that I am lost lol

Answer 4

Okay, so, you also should note that...
\[\huge a^{\frac{m}n}= \sqrt[n]{a^m}\]

Answer 5

= \[\huge [\sqrt[n]a]^m\]

Answer 6

So that makes it...
\[\huge = \frac1{[\color{red}{\sqrt[3]8}]^2}\]

Answer 7

Ok I am confused with = [a − − √ n ] m
why did you do that?

Answer 8

No, that's kinda the definition of a fractional exponent...
\[\huge a^{\frac{m}n}= \sqrt[n]{a^m}=[\sqrt[n]a]^m\]
It becomes a radical, with the index being the denominator, and the entire thing is raised to the numerator...

Answer 9

<ehem>
*index being the denominator*
\[\huge a^{\frac{m}{\color{red}n}}= \sqrt[\color{red}n]{a^m}=[\sqrt[\color{red}n]a]^m\]


*and the entire thing is raised to the numerator*
\[\huge a^{\frac{\color{blue}m}n}= \sqrt[n]{a^{\color{blue}m}}=[\sqrt[n]a]^{\color{blue}m}\]

Answer 10

so it would be (3/8)-2 (I have no idea how to add the brackets im sorry)

Answer 11

just stick to this:\[\huge = \frac1{[\color{red}{\sqrt[3]8}]^2}\]

Answer 12

also there is an answer key to this which is
a) 1/4 b)4 c) 2 |/3 d ) 16 e) -1/4

Answer 13

no need for that, what is the cube root of 8?

Answer 14

c has that weird dividion sigh I forgot the name for it

Answer 15

square root sign
or radical sign

Answer 16

64

Answer 17

I think it is a radical

Answer 18

I said cube root of 8 :P

Answer 19

ohh 512

Answer 20

Still no...
I said cube ROOT not cube.
Means which number, when cubed, gives you 8?

Answer 21

2

Answer 22

Good :)
So...\[\large \sqrt[3]8 = 2\]


Consequently, our expression becomes...
\[\huge = \frac1{[\color{red}{2}]^2}\]
Care to finish it off? :P

Answer 23

wait what about the negative 2?

Answer 24

There is no negative 2?

Answer 25

yes there is the problem is 8 -2/3

Answer 26

We already got rid of that negative by bringing the 8 to the denominator remember?
Look at the first thing I said ^^

Answer 27

And you said you got that part -.-

Answer 28

lol I just need to try to remember that's all and how come the 2- becomes a positive if I bring it down?

Answer 29

-2*

Answer 30

Yes, it does.
Exponents change sign when you switch them from numerators and denominators...

Answer 31

and numerators and denominators are? (im sorry im a bit rusty when it comes to math)

Answer 32

\[\huge \frac{numerator}{denominator}\]

Answer 33

so if the problem was 8 2/3 would I have to change it to a negative when I switch them?

Answer 34

Yes, if ever you want to bring that 8^(2/3) to the denominator, the exponent becomes negative, though I see no reason to do that, we generally switch from numerator to denominator to get rid of negative exponents and not the other way around...

Answer 35

Well okay thank you very much for the help terenzreignz :o)

Answer 36

no problem :)

Answer 37

And it's Terence btw :P

Answer 38

Well thank you Terence and my name is Kay :0)

Answer 39

Noted :)