Question

How do you evaluate `( \frac { 8} { 27} ) ^ { - \frac { 2} { 3} }` without using a calculator?

Answer 1

`(8/27)^(-2/3)=9/4`

Explanation:

As `x^(-1)=1/x`, `x^(1/n)=root(n)x` and `x^(mn)=(x^m)^n`

`(8/27)^(-2/3)`

= `((8/27)^(-1))^(2/3)`

= `(1/(8/27))^(2/3)`

= `((27/8)^(1/3))^2`

= `(root(3)27/root(3)8)^2`

= `(root(3)(3xx3xx3)/root(3)(2xx2xx2))^2`

= `(3/2)^2`

= `9/4`

Answer 2

See a solution process below:

Explanation:

First, rewrite this expression as:

`(8/27)^(-2/3) => (8/27)^(1/3 xx -2)`

We can rewrite this using this rule for exponents:

`x^(color(red)(a) xx color(blue)(b)) = (x^color(red)(a))^color(blue)(b)`

`(8/27)^(color(red)(1/3) xx color(blue)(-2)) => ((8/27)^(color(red)(1/3)))^ color(blue)(-2)`

We can now write this in radical form using this rule:

`x^(1/color(red)(n)) = root(color(red)(n))(x)`

`((8/27)^(1/color(red)(3)))^-2 = (root(color(red)(3))(8/27))^-2`

We can now use this rule for dividing radicals to evaluate the radical:

`root(n)(color(red)(a)/color(blue)(b)) = root(n)(color(red)(a))/root(n)(color(blue)(b))`

`(root(3)(color(red)(8)/color(blue)(27)))^-2 => (root(3)(color(red)(8))/root(3)(color(blue)(27)))^-2 => (root(3)(color(red)(2 * 2 * 2))/root(3)(color(blue)(3 * 3 * 3)))^-2 => (2/3)^-2`

Yet again, we can rewrite this as:

`(2/3)^-2 => (2 xx 1/3)^-2 => 2^-2 xx 1/3^-2`

Now, using these rules of exponents, yes, we can rewrite this again and then evaluate:

`x^color(red)(a) = 1/x^color(red)(-a)` and `1/x^color(blue)(a) = x^color(blue)(-a)`

`2^color(red)(-2) xx 1/3^color(blue)(-2) => 1/2^color(red)(- -2) xx 3^color(blue)(- -2) => 1/2^color(red)(2) xx 3^color(blue)(2) => 1/4 xx 9 =>`

`9/4`