Homework Help

`((-2)(a^-0.5)(b))^4/(3(c^-2)(b^-1))*(b^2/2a)`  Simplify the expression,giving your...

user profile pic

phanpal999 | Student, Grade 8 | (Level 1) Salutatorian

Posted April 5, 2013 at 1:34 PM via web

dislike 2 like


Simplify the expression,giving your answer in positive indices.

Please explain how you arrive at the answer.Thanks

1 Answer | Add Yours

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted April 5, 2013 at 3:36 PM (Answer #1)

dislike 1 like

You need to use the order of operations, hence, you need to raise to the 4th power the product of factors `(-2*a^(-0.5)*b)` , such that:

`(-2*a^(-0.5)*b)^4 = (-2)^4*(a^(-0.5*4))*b^4`

`(-2*a^(-0.5)*b)^4 = 16*(a^(-2))*b^4`

Replacing `16*(a^(-2))*b^4` for `(-2*a^(-0.5)*b)^4` yields:


You need to use the following exponential laws, such that:

`x^(-alpha) = 1/(x^(alpha))`

`x^(alpha)*x^(beta) = x^(alpha + beta)`

`x^(alpha)/x^(beta) = x^(alpha - beta)`

Reasoning by analogy, yields:

`(16*(a^(-2))*b^4*b^2*a)/(3*c^(-2)*b^(-1)*2) = (16/(2*3))*a^(-2+1)*b^(4+2+1)*c^2`

Reducing duplicate factors yields:

`(16*(a^(-2))*b^4*b^2*a)/(3*c^(-2)*b^(-1)*2) = (8/3)*a^(-1)*b^7*c^2`

Hence, performing the simplifications, yields `(16*(a^(-2))*b^4*b^2*a)/(3*c^(-2)*b^(-1)*2) = (8/3)*(b^7*c^2)/a.`

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes