How do you simplify #(g^6h^2m)/(hg^7)#?

Question

2078 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-17T23:57:56

#(hm)/g#

#(g^6h^2m)/hg^7 = g^6/g^7*h^2/h*m = 1/g*h*m=(hm)/g#

Answer 2 · 2017-04-17T23:58:09

See the entire solution process below:

First, use this rule of exponents to modify the #h# term in the denominator:

#a = a^color(blue)(1)#

#(g^6h^2m)/(hg^7) = (g^6h^2m)/(h^color(blue)(1)g^7)#

Next, use these rules of exponents to simplify the #g# and #h# terms in the numerator and denominator:

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

#(g^6h^2m)/(h^color(red)(1)g^7) = (h^(color(red)(2)-color(blue)(1))m)/g^(color(blue)(7)-color(red)(6)) = (h^1m)/g^1#

Now, we can use the reverse of the first rule to finalize the simplification:

#a^color(red)(1) = a#

#(h^1 m)/(g^1) = (hm)/g#