Question

Which expression is equivalent to (startfraction (3 x y superscript negative 5 baseline) cubed over (x superscript negative 2 baseline y squared) superscript negative 4 baseline) endfraction) superscript negative 2? assume startfraction x superscript 10 baseline y superscript 14 baseline over 729 endfraction.

Answer 1

The equivalent expression of $(\frac{(3xy^{-5})^3}{(x^{-2} y^2) ^{-4}})^{-2}$ is $\frac{x^{10} y^{14}}{729}$

How to determine the equivalent expression?

The expression is given as:

$(\frac{(3xy^{-5})^3}{(x^{-2} y^2) ^{-4}})^{-2$

Expand the expression

$(\frac{(3xy^{-5})^3}{(x^{-2} y^2) ^{-4}})^{-2} = (\frac{27x^3y^{-15}}{(x^{8} y^{-8}})^{-2$

Apply the law of indices

$(\frac{(3xy^{-5})^3}{(x^{-2} y^2) ^{-4}})^{-2} = (\frac{27}{(x^{5} y^{7}})^{-2$

Take the inverse of the expression

$(\frac{(3xy^{-5})^3}{(x^{-2} y^2) ^{-4}})^{-2} = (\frac{x^{5} y^{7}}{27})^2$

Apply the square exponent to the expression

$(\frac{(3xy^{-5})^3}{(x^{-2} y^2) ^{-4}})^{-2} = \frac{x^{10} y^{14}}{729}$

Hence, the equivalent expression of $(\frac{(3xy^{-5})^3}{(x^{-2} y^2) ^{-4}})^{-2}$ is $\frac{x^{10} y^{14}}{729}$

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