How do you simplify #\frac { ( - 8) ^ { 4} \cdot 16^ { - 3} \cdot 35^ { 3} } { 14^ { 3} \cdot 50^ { 2} \cdot 24^ { - 2} }#?

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Answer 1 · 2018-05-26T04:04:51

#18/5 =3.6#

#((-8)^4*16^-3*35^3)/(14^3*50^2*24^-2#

=#(-8)^4*16^-3*35^3*14^-3*50^-2*24^2#

=#((-2)^3)^4*(2^4)^-3*5^3*7^3*2^-3*7^-3*2^-2*(5^2)^-2*3^2*(2^3)^2#

= #(-2)^12*2^-12*2^(-3-2+6)*3^2*5^(3-4)*7^(3-3)#

=#2^12*2^-12*2^1*3^2*5^-1*7^0#

= #(2*9*1)/5#

=#18/5#

=#3.6#

Note:
# 1." "(-2)^12 = 2^12# since exponent is even
# 2." "7^0 = 1# .... by definition

Answer 2 · 2018-05-26T09:02:24

#18/5#

#\frac { ( - 8) ^ { 4} \cdot 16^ { - 3} \cdot 35^ { 3} } { 14^ { 3} \cdot 50^ { 2} \cdot 24^ { - 2} }#

Write the bases as the product of their prime factors:

#((-2^3)^4 * (2^4)^-3*(5xx7)^3)/((2xx7)^3*(2xx5^2)^2* (2^3xx3)^-2#

Multiply the indices to remove the brackets.

#=(2^12 * 2^-12 * 5^3 * 7^3)/(2^3 * 7^3 * 2^2 * 5^4 * 2^-6 * 3^-2)#

Simplify by adding the indices of like bases

#= (2^0 * 5^3 * 7^3)/(2^-1 * 3^-2 * 5^4* 7^3)" "larr (7^3/7^3 =1)#

#=(2 * 3^2)/5" "larr" "# law: #1/x^-m = x^m#

#=18/5#