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

3 Answers
Jul 7, 2017

Answer:

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)#

Jul 7, 2017

Answer:

#(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) #

Jul 7, 2017

Answer:

#(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)#