How do you evaluate $3^{4} \cdot 12^{\frac{3}{2}} \cdot {(\frac{1}{2^{2}})}^{\frac{3}{2}}$ $3^412^(3/2)(1/2^2)^(3/2)$ ?

Question

How do you evaluate $3^{4} \cdot 12^{\frac{3}{2}} \cdot {(\frac{1}{2^{2}})}^{\frac{3}{2}}$ $3^412^(3/2)(1/2^2)^(3/2)$ ?

2 Answers

Impact of this question

3630 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-30T07:17:30

$3^{4} \cdot 12^{\frac{3}{2}} \cdot {(\frac{1}{2^{2}})}^{\frac{3}{2}} = 243 \sqrt{3}$ $3^4*12^(3/2)*(1/2^2)^(3/2)=243sqrt3$

Explanation:

$3^{4} \cdot 12^{\frac{3}{2}} \cdot {(\frac{1}{2^{2}})}^{\frac{3}{2}}$ $3^4*12^(3/2)*(1/2^2)^(3/2)$

= $3^{4} \times {(2 \times 2 \times 3)}^{\frac{3}{2}} \times {(2^{- 2})}^{\frac{3}{2}}$ $3^4xx(2xx2xx3)^(3/2)xx(2^(-2))^(3/2)$

= $3^{4} \times {(2 \times 2 \times 3)}^{\frac{3}{2}} \times (2^{(- 2) \cdot \frac{3}{2}})$ $3^4xx(2xx2xx3)^(3/2)xx(2^((-2)*3/2))$

= $3^{4} \times {(2^{2} \times 3)}^{\frac{3}{2}} \times (2^{- 3})$ $3^4xx(2^2xx3)^(3/2)xx(2^(-3))$

= $3^{4} \times {(2^{2})}^{\frac{3}{2}} \times 3^{\frac{3}{2}} \times (2^{- 3})$ $3^4xx(2^2)^(3/2)xx3^(3/2)xx(2^(-3))$

= $3^{4} \times 2^{2 \times \frac{3}{2}} \times 3^{\frac{3}{2}} \times 2^{- 3}$ $3^4xx2^(2xx3/2)xx3^(3/2)xx2^(-3)$

= $3^{4} \times 2^{3} \times 3^{\frac{3}{2}} \times 2^{- 3}$ $3^4xx2^3xx3^(3/2)xx2^(-3)$

= $3^{4 + \frac{3}{2}} \times 2^{3 - 3}$ $3^(4+3/2)xx2^(3-3)$

= $3^{5 + \frac{1}{2}} \times 2^{0}$ $3^(5+1/2)xx2^0$

= $3^{5} \sqrt{3} \times 1 = 243 \sqrt{3}$ $3^5sqrt3xx1=243sqrt3$

Answer 2 · 2016-05-01T19:28:50

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:
= $3^{4} \times {(2^{2} \times 3)}^{\frac{3}{2}} \times \frac{1}{{(2^{2})}^{\frac{3}{2}}}$ $3^4xx(2^2xx3)^(3/2)xx1/(2^2)^(3/2)$

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

= $3^{4} \times (2^{3}) \times 3^{\frac{3}{2}} \times \frac{1}{2^{3}}$ $3^4xx(2^3) xx 3^(3/2) xx 1/(2^(3))$

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.

Science

Math

Humanities

... and beyond

How do you evaluate $3^{4} \cdot 12^{\frac{3}{2}} \cdot {(\frac{1}{2^{2}})}^{\frac{3}{2}}$ $3^412^(3/2)(1/2^2)^(3/2)$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you evaluate 3^4*12^(3/2)*(1/2^2)^(3/2) 34⋅1232⋅(122)32 3^4*12^(3/2)*(1/2^2)^(3/2) ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you evaluate $3^{4} \cdot 12^{\frac{3}{2}} \cdot {(\frac{1}{2^{2}})}^{\frac{3}{2}}$ $3^412^(3/2)(1/2^2)^(3/2)$ ?