How do you simplify #((m^3p^5)/n^7)^6*((m^2n^0p^3)/(m^4n^2))^3# and write it using only positive exponents?

1 Answer
Mar 19, 2017

Answer:

#(m^12color(white)(.)p^(39))/n^(48)#

Explanation:

#color(blue)("Preamble")#

Consider the context of #n^0#
#" "#suppose we had #2^3/2^3 =1#

#" "#This can be written as #2^(3-3) = 2^0=1#

#" "# so #n^0# disappears in the multiplication
....................................................................................
Consider the context of #((m^3p^5)/n^7)^6#

Suppose we had #p^2# then this is #pxxp#

So whatever #p# is it is multiplied by itself twice.

Set #p=3^2# then we have #3^2xx3^2 = 3xx3xx3xx3 = 3^4#

So #(3^2)^2=3^(2xx2)=3^4#
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#color(blue)("Answering the question - applying the above principles")#

Given:#" "((m^3p^5)/n^7)^6xx((m^2n^0p^3)/(m^4n^2))^3#

#=(m^(18)p^(30))/n^42 xx (m^6p^9)/(m^(12)n^6)#

#=(m^(18+6)color(white)(.)p^(30+9))/(m^12color(white)(.)n^(42+6))" "# this is the same as: #" "(m^12xxcancel(m^12))/(cancel(m^12)) xx(p^39)/(n^(48)#

#=(m^12color(white)(.)p^(39))/n^(48)#