Question

How do you simplify `16^(-2/3)`?

Answer 1

`color(blue)(16^(-2/3) = 1/(4*2^(2/3`

Explanation:

Given:
`" "`
`color(blue)(16^(-2/3)`

Identities used:

`color(red)(a^(-b)=1/(a^b)`

`color(red)(a^(m/n) = rootn(a^m)`

Consider the expression given:

`color(blue)(16^(-2/3)`

`rArr 1/(16^(2/3))`

`rArr 1/root3(16^2)`

`rArr 1/root3(256`

`rArr 1/root3(64*4)`

Note that `color(blue)(64 = 4^3`

`rArr 1/root3(4^3*4)`

`rArr 1/root3(4^3)*1/root3(4)`

Note that `color(red)(rootp(m^n) = (m^n)^(1/p)`

`rArr 1/(4^3)^(1/3)*1/root3(4)`

`rArr (1/4)*1/root3(4)`

`rArr (1/4)*1/root3(2^2)`

Note that `color(red)(sqrt(m^n) = (m^n)^(1/2)=m^(n/2)`

`rArr 1/4*[1/[(2^2)^(1/3))]`

`rArr 1/4*1/(2^(2/3))`

`rArr 1/(4*2^(2/3)`

Hence,

`color(blue)(16^(-2/3) = 1/(4*2^(2/3)`

If you wish, you can continue to simplify further:

`1/(4*2^(2/3)`

`rArr 1/(2^2*2^(2/3)`

Note that `color(red)(a^m*a^n=a^(m+n)`

`rArr 1/2^(2+(2/3)`

`rArr 1/(2^(8/3)`

`rArr 2^(-8/3`

Hope you find this solution useful.