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

Apr 30, 2016

${3}^{4} \cdot {12}^{\frac{3}{2}} \cdot {\left(\frac{1}{2} ^ 2\right)}^{\frac{3}{2}} = 243 \sqrt{3}$

#### Explanation:

${3}^{4} \cdot {12}^{\frac{3}{2}} \cdot {\left(\frac{1}{2} ^ 2\right)}^{\frac{3}{2}}$

= ${3}^{4} \times {\left(2 \times 2 \times 3\right)}^{\frac{3}{2}} \times {\left({2}^{- 2}\right)}^{\frac{3}{2}}$

= ${3}^{4} \times {\left(2 \times 2 \times 3\right)}^{\frac{3}{2}} \times \left({2}^{\left(- 2\right) \cdot \frac{3}{2}}\right)$

= ${3}^{4} \times {\left({2}^{2} \times 3\right)}^{\frac{3}{2}} \times \left({2}^{- 3}\right)$

= ${3}^{4} \times {\left({2}^{2}\right)}^{\frac{3}{2}} \times {3}^{\frac{3}{2}} \times \left({2}^{- 3}\right)$

= ${3}^{4} \times {2}^{2 \times \frac{3}{2}} \times {3}^{\frac{3}{2}} \times {2}^{- 3}$

= ${3}^{4} \times {2}^{3} \times {3}^{\frac{3}{2}} \times {2}^{- 3}$

= ${3}^{4 + \frac{3}{2}} \times {2}^{3 - 3}$

= ${3}^{5 + \frac{1}{2}} \times {2}^{0}$

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

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:
= ${3}^{4} \times {\left({2}^{2} \times 3\right)}^{\frac{3}{2}} \times \frac{1}{{2}^{2}} ^ \left(\frac{3}{2}\right)$

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

= ${3}^{4} \times \left({2}^{3}\right) \times {3}^{\frac{3}{2}} \times \frac{1}{{2}^{3}}$

Add the indices of like bases and cancel like factors

= ${3}^{4 + \frac{3}{2}} \times \cancel{{2}^{3}} \times \frac{1}{\cancel{{2}^{3}}}$

= ${3}^{4 + 1 \frac{1}{2}}$

= ${3}^{5 \frac{1}{2}}$ = ${3}^{\frac{11}{2}}$

= ${\left(\sqrt{3}\right)}^{11}$

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