Simplifying (32x^5)^-2/5

Hi,

I'm trying to simplify (32x^5)^-2/5

I have the written solutions, which says it then goes to (2x)^-2, and then to 1/(4x^2)

I get how (2x)^-2 goes to 1/(4x^2), but how does (32x^5)^-2/5 go to (2x)^-2.

Please can you explain that one step to me?

Thanks

Reply 1

blobbybill

OP

11

Original post by AnIndianGuy

1/(4x^2) is equivalent to (2x)^-2. Both are correct.

I know. I am asking how you simplify (32x^5)^-2/5 into 2x^-2

Reply 2

Notnek

21

Original post by blobbybill

Hi,

I'm trying to simplify (32x^5)^-2/5

I have the written solutions, which says it then goes to (2x)^-2, and then to 1/(4x^2)

I get how (2x)^-2 goes to 1/(4x^2), but how does (32x^5)^-2/5 go to (2x)^-2.

Please can you explain that one step to me?

Thanks

(32x^5)^{-\frac{2}{5}} = \left((32x^5)^{\frac{1}{5}} \right) ^{-2} = ...

Does that help?

Reply 3

RogerOxon

21

Original post by blobbybill

how does (32x^5)^-2/5 go to (2x)^-2.

Apply the following to your expression:
(a^b)^c = a^(bc)

You should get 32^d, where d is for you to calculate. Note that you can write 32 as a power of 2.

Reply 4

TheDanHall

1

thanks for the help g

Reply 5

OllyTolly

1

i dont understand lol

Reply 6

MDMA METH

3

(32x^5)^-2/5 first you want to divide the alike powers which in this scenario is 5 now you'll have 32^-2/5 . x^-2 working the indices of 32^-2/5 X 1/(5th root of 32)^2 X indices of x^-2 = 1/x^21/2^2 X 1/X^2 = 1/4 X 1/X^2 = 1/4X^2.I HOPE THIS HELPS

(edited 3 years ago)

Reply 7

davros

16

Original post by MDMA METH

(32x^5)^-2/5 first you want to divide the alike powers which in this scenario is 5 now you'll have 32^-2/5 . x^-2 working the indices of 32^-2/5 X 1/(5th root of 32)^2 X indices of x^-2 = 1/x^21/2^2 X 1/X^2 = 1/4 X 1/X^2 = 1/4X^2.I HOPE THIS HELPS

You're a year too late :biggrin:

Simplifying (32x^5)^-2/5

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you