How do you simplify $- 3 x^{2} y^{3} z \sqrt{75 x^{9} y^{4} z^{10}}$ $-3x^2y^3zsqrt(75x^9y^4z^10)$ ?

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How do you simplify $- 3 x^{2} y^{3} z \sqrt{75 x^{9} y^{4} z^{10}}$ $-3x^2y^3zsqrt(75x^9y^4z^10)$ ?

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Answer 1 · 2015-06-01T01:49:56

Remembering one exponential law which states that $a^{\frac{n}{m}} = \sqrt[m]{a^{n}}$ $a^(n/m)=root(m)(a^n)$ , we can rewrite the function as

$- 3 x^{2} y^{3} z 75^{\frac{1}{2}} x^{\frac{9}{2}} y^{\frac{4}{2}} z^{\frac{10}{2}}$ $-3x^2y^3z75^(1/2)x^(9/2)y^(4/2)z^(10/2)$

(As this is a square root, we have that, for the exponential rule, $m = 2$ $m=2$ )

Now, following another law of exponentials, we have that $a^{n} \cdot a^{m} = a^{n + m}$ $a^n*a^m=a^(n+m)$

$- 3 x^{(2 + \frac{9}{2})} y^{(3 + 2)} z^{(1 + \frac{10}{2})} 75^{\frac{1}{2}}$ $-3x^((2+9/2))y^((3+2))z^((1+10/2))75^(1/2)$

$- 3 x^{\frac{13}{2}} y^{5} z^{6} 75^{\frac{1}{2}}$ $-3x^(13/2)y^5z^(6)75^(1/2)$

We can rewrite the element $75^{\frac{1}{2}}$ $75^(1/2)$ as ${(3 \cdot 25)}^{\frac{1}{2}} = 3^{\frac{1}{2}} \cdot 25^{\frac{1}{2}}$ $(3*25)^(1/2)=3^(1/2)*25^(1/2)$

Thus, we can combine the first $- 3$ $-3$ with the just-factored $3$ $3$ :

$- {(3)}^{1} \cdot (3^{\frac{1}{2}}) x^{\frac{13}{2}} y^{5} z^{6} 75^{\frac{1}{2}} 25^{\frac{1}{2}}$ $-(3)^(1)*(3^(1/2))x^(13/2)y^5z^(6)75^(1/2)25^(1/2)$

As $25^{\frac{1}{2}} = \sqrt{25} = 5$ $25^(1/2)=sqrt(25)=5$ ,

$5 {(- 3)}^{1.5} x^{6.5} y^{5} z^{6}$ $5(-3)^(1.5)x^(6.5)y^5z^(6)$

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How do you simplify $- 3 x^{2} y^{3} z \sqrt{75 x^{9} y^{4} z^{10}}$ $-3x^2y^3zsqrt(75x^9y^4z^10)$ ?

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