Question

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

Answer 1

`18/5 =3.6`

Explanation:

`((-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

`18/5`

Explanation:

`\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`