`((-2)(a^-0.5)(b))^4/(3(c^-2)(b^-1))*(b^2/2a)`  Simplify the expression,giving your answer in positive indices. Please explain how you arrive at the answer.Thanks

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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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.`

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