Question

How do you simplify `((2a)/(9c))^-2` and write it using only positive exponents?

Answer 1

See a solution process below:

Explanation:

First, use these rules of exponents to eliminate the outer exponent:

`a = a^color(red)(1)` and `(x^color(red)(a))^color(blue)(b) = x^(color(red)(a) xx color(blue)(b))`

`((2a)/(9c))^-2 => ((2^color(red)(1)a^color(red)(1))/(9^color(red)(1)c^color(red)(1)))^color(blue)(-2) => (2^(color(red)(1) xx color(blue)(-2))a^(color(red)(1) xx color(blue)(-2)))/(9^(color(red)(1) xx color(blue)(-2))c^(color(red)(1) xx color(blue)(-2))) => (2^-2a^-2)/(9^-2c^-2)`

Now, use these rules of exponents to eliminate the negative exponents:

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

`(2^color(red)(-2)a^color(red)(-2))/(9^color(red)(-2)c^color(red)( -2)) => (9^color(red)(- -2)c^color(red)(- -2))/(2^color(red)(- -2)a^color(red)(- -2)) => (9^2c^2)/(2^2a^2) => (81c^2)/(4a^2)`

Answer 2

`(81c^2)/(4a^2)`

Explanation:

`((2a)/(9c))^-2`

`;.=1/(((2a)/(9c))^2)`

`;.=1/((4a^2)/(81c^2))`

`;.=1/1 xx (81c^2)/(4a^2)`

`;.= (81c^2)/(4a^2)`

Answer 3

`(81c^2)/(4a^2)`

Explanation:

One of the laws of indices states that:

`(a/b)^-m = (b/a)^(+m)`

So, by inverting the fraction, the index changes sign.

`((2a)/(9c))^color(magenta)(-2) = ((9c)/(2a))^color(magenta)(2)`

All the indices are now positive and it only remains to square each factor.

`((9c)/(2a))^2 = (81c^2)/(4a^2)`