How do you simplify ${(5 x^{2} y^{3})}^{2} \cdot {(2 x^{3} y^{4})}^{3}$ $(5x^2y^3)^2*(2x^3y^4)^3$ and write it using only positive exponents?

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How do you simplify ${(5 x^{2} y^{3})}^{2} \cdot {(2 x^{3} y^{4})}^{3}$ $(5x^2y^3)^2*(2x^3y^4)^3$ and write it using only positive exponents?

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Answer 1 · 2018-01-02T13:44:30

$200 x^{13} y^{18}$ $200x^(13)y^(18)$

Explanation:

$by appling the laws of exponents$ $"by appling the "color(blue)"laws of exponents"$

$∙ x {(a^{m})}^{n} = a^{(m \times n)} \leftarrow (1)$ $•color(white)(x)(a^m)^(n)=a^((mxxn))larr(color(red)(1))$

$∙ x a^{m} \times a^{n} = a^{(m + n)}$ $•color(white)(x)a^mxxa^n=a^((m+n))$

$(1) is extended to include all factors inside the parenthesis$ $(color(red)(1))" is extended to include all factors inside the parenthesis"$

$\Rightarrow {(5 x^{2} y^{3})}^{2} = 5^{(1 \times 2)} \times x^{(2 \times 2)} \times y^{(3 \times 2)}$ $rArr(5x^2y^3)^2=5^((1xx2))xx x^((2xx2))xxy^((3xx2))$

$\times \times \times \times = 5^{2} \times x^{4} \times y^{6} = 25 x^{4} y^{6}$ $color(white)(xxxxxxxx)=5^2xx x^4xx y^6=25x^4y^6$

$\Rightarrow {(2 x^{3} y^{4})}^{3} = 2^{(1 \times 3)} \times x^{(3 \times 3)} \times y^{(4 \times 3)}$ $rArr(2x^3y^4)^3=2^((1xx3))xxx^((3xx3))xxy^((4xx3))$

$\times \times \times \times = 2^{3} \times x^{9} \times y^{12} = 8 x^{9} y^{12}$ $color(white)(xxxxxxxx)=2^3xx x^9xxy^(12)=8x^9y^(12)$

$\Rightarrow {(5 x^{2} y^{3})}^{2} \times {(2 x^{3} y^{4})}^{3}$ $rArr(5x^2y^3)^2xx(2x^3y^4)^3$

$= 25 x^{4} y^{6} \times 8 x^{9} y^{12}$ $=25x^4y^6xx8x^9y^(12)$

$= (25 \times 8) \times (x^{4} \times x^{9}) \times (y^{6} \times y^{12})$ $=(25xx8)xx(x^4xx x^9)xx(y^6xxy^(12))$

$= 200 \times x^{(4 + 9)} \times y^{(6 + 12)}$ $=200xx x^((4+9))xxy^((6+12))$

$= 200 x^{13} y^{18}$ $=200x^(13)y^(18)$

Answer 2 · 2018-01-02T13:46:32

See a solution process below:

Explanation:

First, use these rules for exponents to eliminate the outer exponents for each term:

or $a = a^{1}$ $a = a^color(red)(1)$ and ${(x^{a})}^{b} = x^{a \times b}$ $(x^color(red)(a))^color(blue)(b) = x^(color(red)(a) xx color(blue)(b))$

${(5 x^{2} y^{3})}^{2} \cdot {(2 x^{3} y^{4})}^{3} \Rightarrow$ $(5x^2y^3)^2 * (2x^3y^4)^3 =>$

${(5^{1} x^{2} y^{3})}^{2} \cdot {(2^{1} x^{3} y^{4})}^{3} \Rightarrow$ $(5^color(red)(1)x^color(red)(2)y^color(red)(3))^color(blue)(2) * (2^color(red)(1)x^color(red)(3)y^color(red)(4))^color(blue)(3) =>$

$(5^{1 \times 2} x^{2 \times 2} y^{3 \times 2}) \cdot (2^{1 \times 3} x^{3 \times 3} y^{4 \times 3}) \Rightarrow$ $(5^(color(red)(1) xx color(blue)(2))x^(color(red)(2) xx color(blue)(2))y^(color(red)(3) xx color(blue)(2))) * (2^(color(red)(1) xx color(blue)(3))x^(color(red)(3) xx color(blue)(3))y^(color(red)(4) xx color(blue)(3))) =>$

$(5^{2} x^{4} y^{6}) \cdot (2^{3} x^{9} y^{12}) \Rightarrow$ $(5^2x^4y^6) * (2^3x^9y^12) =>$

$(25 x^{4} y^{6}) \cdot (8 x^{9} y^{12})$ $(25x^4y^6) * (8x^9y^12)$

Next, rewrite the expression as:

$(25 \cdot 8) (x^{4} \cdot x^{9}) (y^{6} \cdot y^{12}) \Rightarrow$ $(25 * 8)(x^4 * x^9)(y^6 * y^12) =>$

$200 (x^{4} \cdot x^{9}) (y^{6} \cdot y^{12})$ $200(x^4 * x^9)(y^6 * y^12)$

Now, use this rule of exponents to complete the simplification:

$x^{a} \times x^{b} = x^{a + b}$ $x^color(red)(a) xx x^color(blue)(b) = x^(color(red)(a) + color(blue)(b))$

$200 (x^{4} \cdot x^{9}) (y^{6} \cdot y^{12}) \Rightarrow$ $200(x^color(red)(4) * x^color(blue)(9))(y^color(red)(6) * y^color(blue)(12)) =>$

$200 x^{4 + 9} y^{6 + 12} \Rightarrow$ $200x^(color(red)(4)+color(blue)(9))y^(color(red)(6)+color(blue)(12)) =>$

$200 x^{13} y^{18}$ $200x^13y^18$

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