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

Question

Kennedy Adams · Accepted Answer

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

According to one of the indices' laws,

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

Thus, the index changes sign when the fraction is inverted.

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

Now that every index is positive, all that needs to be done is square each factor.

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

Savannah Anderson · Answer

See a solution process below:

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)#

Genesis Allen · Answer

#(81c^2)/(4a^2)# #((2a)/(9c))^-2# #;.=1/(((2a)/(9c))^2)# #;.=1/((4a^2)/(81c^2))# #;.=1/1 xx (81c^2)/(4a^2) # #;.= (81c^2)/(4a^2) #

Isaiah Anthony · Answer

undefined To simplify ((2a)/(9c))^-2 and write it using only positive exponents, you apply the negative exponent rule, which states that a^(-n) = 1/(a^n). So, ((2a)/(9c))^-2 becomes (9c/(2a))^2.