Question

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

Answer 1

`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`

Answer 2

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.