Question

Please please help ASAP I thought I knew this but..nope.

(2^2/3•3^2/3)^-6

Also I want to check my answer for this problem 8^-1/2

Thanks so much.

Answer 1

so the problem is really

\[(\frac{2^2}{3} \times \frac{3^2}{3})^{-6}\]

which becomes

\[(\frac{2^2 \times 3^2}{3 \times 3})^{-6} \]

when dividing the same base subtract the powers

\[(2^2 \times 3^{2 - 1 - 1})^{-6} = (2^2 \times 3^0)^{-6}\]

so you now need to know about power of zero being 1

so the problem is really

\[(2^2)^{-6}\]

to simplify this, use power of a power, and multiply the powers.

then you may need to recognise the negative power indicates a fraction.

hope it helps