Question

Solve the equation with rational exponents.

Answer 1

(x-6) ^3/2 =27

Answer 2

I dont remember how to do this..

Answer 3

\[\large (x-6) ^{\frac{3}{2}} =27\]

yes?

Answer 4

Yeah

Answer 5

because \(\large n^\frac{a}{b} = n^{\frac{a}{1}\times \frac{1}{b}} = (n^a)^{\frac{1}{b}} = (n^{\frac{1}{b}})^a\)

Answer 6

Um okay?

Answer 7

or if its simpler

https://gyazo.com/dfe8c6b9853179633b8a8fe3a0bcb746

Answer 8

so in \[\frac{3}{2}\] you have a cube and a square root

you want to undo them both, but one by one

Answer 9

\[n^{3/2} = (n^{1/2})^3 = (n^3)^{1/2}\]

Answer 10

and here \(n = x-6\) !!

Answer 11

Um all im really understanding is that n= x-6 ... Except I dont get the point of n

Answer 12

it is an attempt to simplify but it might just be doing the opposite

you know that \[n^{1/2} = \sqrt{n}\] yes?

Answer 13

Yes

Answer 14

and \[n^{3/2} = (n^{1/2})^3\]

yes

Answer 15

That I dont remember

Answer 16

I thought im suppose to flip 3/2 to 2/3 to cancel it out and put 2/3 by the 27...

Answer 17

if that's how you see it, go for it. it is how you solve it.

Answer 18

Okay. But I dont remember what to do after that

Answer 19

well, that is what i am trying to get at. you now have \[x - 6 = 27^{\frac{2}{3}}\]

but do you know what that 2/3 actually means?

Answer 20

Um its like the reciprocal of 3/2

Answer 21

\[n^{2/3} = (n^2)^{1/3} = \sqrt[3]{n^2}\]
and
\[n^{2/3} = (n^{1/3})^{2} = (\sqrt[3]{n})^2\]

Answer 22

Okay so your saying...

x-6= ^3 square root 27 ^2

Answer 23

sorry, don't really get that notation, just feel free to post what you think the answer is