How do you simplify #(12x^(2)y^(-2))^(5)(4xy^(-3))^(-8)#?

1 Answer
Oct 16, 2015

Answer:

#(81x^2y^14)/(64) #

Very detailed answer. The more mathematically able can skip many parts.

Explanation:

Look at the two parts separately then put them together afterwards
and simplify. As a side note to aid understanding: if we were to have general form of #z^-2# this may be written as #1/(z^2)#

Consider the first part: #(12x^2y^(-2))^5#

Rewrite the bit inside the brackets as # (12x^2)/y^2#

So the whole part may be rewritten as:

#( (12x^2)/(y^2))^5#

A method example: # (x^2)^3# is the same as #x^2 times x^2 times x^2 =x^(2 times 3) = x^6#

Back to the question:

#( (12x^2)/(y^2))^5 =(12^5x^(2 times 5))/(y^(2 times 5)) = (12^5x^10)/y^10# ......... ( 1 )

Consider the second part: #(4xy^(-3))^(-8)#

Rewrite the bit inside the bracket as #(4x)/y^(3)#

So the whole part is may be written as:

#(4x/y^(3))^(-8)#

Although it is not normally written like this it gives:
#1/((4x)/(y^3))^8#

The net result of this is:

#((y^3)/(4x))^8 = (y^24)/(4^8x^8)# ....... ( 2 )

Putting (1) and (2) together gives:

#(12^5x^10)/y^10 times (y^24)/(4^8x^8)#

Grouping variables and constants we have:

( remember that #12^5 = (3 times 4)^5 = 3^5 times 4^5#)

so we have:

#(3^4 times 4^5)/4^8 times x^10/x^8 times (y^24)/(y^10) #

Simplifying -> #81/64 times x^2 times y^14#

#(81x^2y^14)/(64) #