What is ((2x^3)/(x^3*2x^2*2x^4))^4 ?

The solution to ((2x^3)/(x^3*2x^2*2x^4))^4 is 1/(16x^{24)}

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simplify ((2x^3)/(x^32x^22x^4))^4

simplify

(\frac{2 x ^{3}}{x ^{3} \cdot 2 x ^{2} \cdot 2 x ^{4}})^{4}

\frac{1}{16 x ^{24}}

Solution steps

(\frac{2 x ^{3}}{x ^{3} \cdot 2 x ^{2} \cdot 2 x ^{4}})^{4}

Apply exponent rule:

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

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

Simplify

(2 x^{3})^{4} : 16 x^{12}

= \frac{16 x ^{12}}{( x ^{3} \cdot 2 x ^{2} \cdot 2 x ^{4} ) ^{4}}

Simplify

(x^{3} \cdot 2 x^{2} \cdot 2 x^{4})^{4} : 256 x^{36}

= \frac{16 x ^{12}}{256 x ^{36}}

Cancel the common factor:

16

= \frac{x ^{12}}{16 x ^{36}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{x ^{12}}{x ^{36}} = \frac{1}{x ^{36 - 12}}

= \frac{1}{16 x ^{36 - 12}}

Subtract the numbers:

36 - 12 = 24

= \frac{1}{16 x ^{24}}

4 x < - 12

x^{3} - 6 x^{2} + 11 x - 6

x^{2} + 2 x - 2 = 0

4 x^{2} - 12 x + 8

\frac{8}{- 2}

What is ((2x^3)/(x^3*2x^2*2x^4))^4 ?
The solution to ((2x^3)/(x^3*2x^2*2x^4))^4 is 1/(16x^{24)}