Question

What is `root4( \frac { 16x ^ { 4} } { 81x ^ { - 8} } )`?

Answer 1

I found: `root(4)((16x^4)/(81x^-8))=2/3x^3`

Explanation:

You can solve it remembering that:
`2^4=2*2*2*2=16`
`3^4=3*3*3*3=81`
and
`x^-8=1/x^8` and also: `1/x^-8=x^8`
so you can write:
`root(4)(x^4/x^-8)=root(4)(x^4x^8)=root(4)(x^(4+8))=root(4)(x^12)`
remembering the fact that a root corresponds to a fractional exponent you get:
`root(4)(x^12)=x^(12*1/4)=x^3`
so at the end your original root will give you:
`root(4)((16x^4)/(81x^-8))=2/3x^3`