How do you evaluate 3^4*12^(3/2)*(1/2^2)^(3/2) ?

2 Answers
Apr 30, 2016

3^4*12^(3/2)*(1/2^2)^(3/2)=243sqrt3

Explanation:

3^4*12^(3/2)*(1/2^2)^(3/2)

= 3^4xx(2xx2xx3)^(3/2)xx(2^(-2))^(3/2)

= 3^4xx(2xx2xx3)^(3/2)xx(2^((-2)*3/2))

= 3^4xx(2^2xx3)^(3/2)xx(2^(-3))

= 3^4xx(2^2)^(3/2)xx3^(3/2)xx(2^(-3))

= 3^4xx2^(2xx3/2)xx3^(3/2)xx2^(-3)

= 3^4xx2^3xx3^(3/2)xx2^(-3)

= 3^(4+3/2)xx2^(3-3)

= 3^(5+1/2)xx2^0

= 3^5sqrt3xx1=243sqrt3

May 1, 2016

Once the concepts have been mastered, the details are shown by the previous contributor, it is important to streamline your working.

Explanation:

Change any base into prime factors:
= 3^4xx(2^2xx3)^(3/2)xx1/(2^2)^(3/2)

Remove brackets by multiplying the indices.
(take care with the multiplication of fractions)

= 3^4xx(2^3) xx 3^(3/2) xx 1/(2^(3))

Add the indices of like bases and cancel like factors

= 3^(4+3/2) xx cancel(2^3)xx 1/(cancel(2^(3)))

= 3^(4+1 1/2)

= 3^(5 1/2) = 3^(11/2)

= (sqrt3)^11

Which form of the answer is best or simplest is debatable.