How do you rewrite #(7ab^-2)/(3x)# using a positive exponent?

1 Answer
Oct 20, 2015

Answer:

#(7ab^(-2))/(3x) = (7a)/(3b^2x)#

Explanation:

#b^3 = b xx b xx b#

#b^2 = b^3/b = b xx b#

#b^1 = b^2/b = b#

#b^0 = b^1/b = 1#

#b^(-1) = b^0/b = 1/b#

#b^(-2) = b^(-1)/b = (1/b)/b = 1/b^2#

And in other news:
#(7ab^(-2))/(3x) = (7a)/(3x) * b^(-2)#

#color(white)("XXXX") = (7a)/(3x)*1/b^2#

#color(white)("XXXX") = (7a)/(3xb^2)#