Question

If k=64, how do you evaluate `1/2k^(2/3)+5(k^(1/2))^(2/3)`?

Answer 1

`28`

Explanation:

Using the `color(blue)"laws of exponents"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(a^(m/n)=(root(n)(a))^m" and "(a^m)^n=a^(mxxn))color(white)(2/2)|)))`

`rArr1/2k^(2/3)+5(k^(1/2))^(2/3)`

`=1/2k^(2/3)+5k^(1/2xx2/3)`

`=1/2k^(2/3)+5k^(1/3)`

[let k = 64]

`=(1/2xx(root(3)(64))^2)+(5xxroot(3)(64))`

`=(1/2xx4^2)+(5xx4)`

`=8+20=28`