Question

Simplify. square root of 9 over sixth root of 9

9 to the power of negative 2 over 3

9 to the power of negative 1 over 3

8 to the power of 3 over 10

9 to the power of 1 over 3

Answer 1

\[\Large \frac{ \sqrt{9} }{ \sqrt[6]{9} }\]correct?

Answer 2

\(\frac{a^b}{a^c}=a^{b-c}=\frac{1}{a^{c-b}}\\\sqrt[a]{b^c}=b^{\frac{c}{a}}\\(\frac{a}{b})^{-c}=(\frac{b}{a})^c\)

Answer 3

keep in mind that \(\Large \bf \cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}= \cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}}\implies a^{-\frac{{\color{blue} n}}{{\color{red} m}}} \)

Answer 4

so for the first one

\(\frac{\sqrt{9}}{\sqrt[6]{9}}=\frac{9^{\frac{1}{2}}}{9^{\frac{1}{6}}}=9^{\frac{1}{2}-\frac{1}{6}}=9^{\frac{2}{6}}=\sqrt[6]{9^2}\)

Answer 5

there must be an echo in here

Answer 6

:)

Answer 7

hehe

Answer 8

Okay, thank you :)