# How do you simplify and write [(x^2/5y^3/4)^5]^-4 (36x^-6y^4)^(1/2) with positive exponents?

Sep 17, 2016

$\frac{{20}^{20} \times 6 {y}^{2}}{x} ^ 3$

#### Explanation:

${\left[{\left({x}^{2} / 5 {y}^{3} / 4\right)}^{5}\right]}^{-} 4 {\left(36 {x}^{-} 6 {y}^{4}\right)}^{\frac{1}{2}}$

Remove the outer brackets by multiplying the indices, to give

=${\left({x}^{2} / 5 {y}^{3} / 4\right)}^{-} 20 \times {36}^{\frac{1}{2}} {x}^{-} 3 {y}^{2}$

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Recall: two laws for negative indices and ${x}^{\frac{1}{2}} = \sqrt{x}$

${\left(\frac{a}{b}\right)}^{-} m = {\left(\frac{b}{a}\right)}^{m} \text{ and } {x}^{-} m = \frac{1}{x} ^ m$

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${\left(\frac{5}{x} ^ 2 \frac{4}{y} ^ 3\right)}^{20} \times \frac{\sqrt{36} \times {y}^{2}}{x} ^ 3$

=${20}^{20} \times 6 {y}^{2} / {x}^{3}$

=$\frac{{20}^{20} \times 6 {y}^{2}}{x} ^ 3$