What is ((x^7y^3)/(x^2y))^4 ?

The solution to ((x^7y^3)/(x^2y))^4 is x^{20}y^8

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simplify ((x^7y^3)/(x^2y))^4

simplify

(\frac{x ^{7} y ^{3}}{x ^{2} y})^{4}

x^{20} y^{8}

Solution steps

(\frac{x ^{7} y ^{3}}{x ^{2} y})^{4}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( x ^{7} y ^{3} ) ^{4}}{( x ^{2} y ) ^{4}}

Simplify

(x^{7} y^{3})^{4} : x^{28} y^{12}

= \frac{x ^{28} y ^{12}}{( x ^{2} y ) ^{4}}

Simplify

(x^{2} y)^{4} : x^{8} y^{4}

= \frac{x ^{28} y ^{12}}{x ^{8} y ^{4}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

\frac{x ^{28}}{x ^{8}} = x^{28 - 8}

= \frac{y ^{12} x ^{28 - 8}}{y ^{4}}

Subtract the numbers:

28 - 8 = 20

= \frac{x ^{20} y ^{12}}{y ^{4}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

\frac{y ^{12}}{y ^{4}} = y^{12 - 4}

= x^{20} y^{12 - 4}

Subtract the numbers:

12 - 4 = 8

= x^{20} y^{8}

x^{4} - (x - 2)^{4}

0 = - \frac{3}{2} x + 3

128 = x^{2}

3 x = - 27

2 x = - 8