How do you simplify ${(9^{\frac{1}{2}} \cdot 9^{\frac{2}{3}})}^{\frac{1}{6}}$ $(9^(1/2) * 9^(2/3))^(1/6)$ ?

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How do you simplify ${(9^{\frac{1}{2}} \cdot 9^{\frac{2}{3}})}^{\frac{1}{6}}$ $(9^(1/2) * 9^(2/3))^(1/6)$ ?

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Answer 1 · 2015-05-22T23:55:10

Two important law of exponentials here:

$(a^{n}) (a^{m}) = a^{n + m}$ $(a^n)(a^m)=a^(n+m)$
${(a^{n})}^{m} = a^{n \cdot m}$ $(a^n)^m=a^(n*m)$

${(9^{\frac{1}{2}} 9^{\frac{2}{3}})}^{\frac{1}{6}}$ $(9^(1/2)9^(2/3))^(1/6)$ = ${(9^{\frac{1}{2} + \frac{2}{3}})}^{\frac{1}{6}}$ $(9^(1/2+2/3))^(1/6)$ = ${(9^{\frac{7}{6}})}^{\frac{1}{6}}$ $(9^(7/6))^(1/6)$ = $9^{(\frac{7}{6}) (\frac{1}{6})}$ $9^((7/6)(1/6))$ = $9^{\frac{7}{36}}$ $color(green)(9^(7/36))$

Answer 2 · 2015-05-23T00:00:18

${(9^{\frac{1}{2}} \cdot 9^{\frac{2}{3}})}^{\frac{1}{6}} = {(9^{(\frac{1}{2} + \frac{2}{3})})}^{\frac{1}{6}}$ $(9^(1/2)*9^(2/3))^(1/6) = (9^((1/2+2/3)))^(1/6)$

$= {(9^{\frac{7}{6}})}^{\frac{1}{6}} = 9^{\frac{7}{6} \cdot \frac{1}{6}} = 9^{\frac{7}{36}} = 9^{\frac{1}{2} \cdot \frac{7}{18}}$ $=(9^(7/6))^(1/6) = 9^(7/6*1/6) = 9^(7/36) = 9^(1/2*7/18)$

$= {(9^{\frac{1}{2}})}^{\frac{7}{18}} = {(\sqrt{9})}^{\frac{7}{18}} = 3^{\frac{7}{18}}$ $= (9^(1/2))^(7/18) = (sqrt(9))^(7/18) = 3^(7/18)$

The identities we use here are:

$x^{a} \cdot x^{b} = x^{a + b}$ $x^a * x^b = x^(a+b)$

$x^{a b} = {(x^{a})}^{b}$ $x^(ab) = (x^a)^b$

$x^{\frac{1}{n}} = \sqrt[n]{x}$ $x^(1/n) = root(n)x$

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How do you simplify ${(9^{\frac{1}{2}} \cdot 9^{\frac{2}{3}})}^{\frac{1}{6}}$ $(9^(1/2) * 9^(2/3))^(1/6)$ ?

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