Ross18876 Ross18876
  • 12-04-2017
  • Mathematics
contestada

Cube root of square root of x raised to the fourth power

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ItsLeviosaaa
ItsLeviosaaa ItsLeviosaaa
  • 12-04-2017
[tex] \sqrt[3]{ \sqrt{x} } ^{4} [/tex]
Answer Link
apologiabiology
apologiabiology apologiabiology
  • 12-04-2017
remember
[tex](x^m)^n=x^{mn}[/tex]
and
[tex] \sqrt[n]{x} =x^ \frac{1}{n} [/tex]
and
[tex]x^{m+n}=(x^m)(x^n)[/tex]

so
[tex]( \sqrt[3]{ \sqrt{x} } )^4[/tex]=
[tex]( \sqrt[3]{ x^ \frac{1}{2} } )^4[/tex]=
[tex]((x^ \frac{1}{2})^ \frac{1}{3} )^4[/tex]=
[tex](x^ \frac{1}{6})^4 [/tex]=
[tex]x^ \frac{4}{6} [/tex]=
[tex]x^ \frac{2}{3} [/tex]=
[tex] \sqrt[3]{x^2} [/tex]
Answer Link

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