How do you simplify $\frac{{(2^{3} ∙ {(2^{2})}^{3})}^{2}}{2}$ $((2^3 •(2^2)^3)^2 )/ 2$ ?

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How do you simplify $\frac{{(2^{3} ∙ {(2^{2})}^{3})}^{2}}{2}$ $((2^3 •(2^2)^3)^2 )/ 2$ ?

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Answer 1 · 2016-07-04T06:47:25

$\frac{{(2^{3} \cdot {(2^{2})}^{3})}^{2}}{2} = 2^{17}$ $(2^3*(2^2)^3)^2/2=2^17$

Explanation:

We can use here three formulas relating to exponents;

$a^{m} \cdot a^{n} = a^{(m + n)}$ $a^m*a^n=a^((m+n))$ , $\frac{a^{m}}{a^{n}} = a^{(m - n)}$ $a^m/a^n=a^((m-n))$ and ${(a^{m})}^{n} = a^{(m \times n)}$ $(a^m)^n=a^((mxxn))$

Hence $\frac{{(2^{3} \cdot {(2^{2})}^{3})}^{2}}{2}$ $(2^3*(2^2)^3)^2/2$

= $\frac{{(2^{3} \cdot 2^{(2 \times 3)})}^{2}}{2}$ $(2^3*2^((2xx3)))^2/2$

= $\frac{{(2^{3} \cdot 2^{6})}^{2}}{2}$ $(2^3*2^6)^2/2$

= $\frac{{(2^{(3 + 6)})}^{2}}{2}$ $(2^((3+6)))^2/2$

= $\frac{{(2^{9})}^{2}}{2^{1}}$ $(2^9)^2/2^1$ (as $a = a^{1}$ $a=a^1$ )

= $\frac{2^{(9 \times 2)}}{2^{1}}$ $2^((9xx2))/2^1$

= $\frac{2^{18}}{2^{1}}$ $2^18/2^1$

= $2^{(18 - 1)}$ $2^((18-1))$

= $2^{17}$ $2^17$

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How do you simplify $\frac{{(2^{3} ∙ {(2^{2})}^{3})}^{2}}{2}$ $((2^3 •(2^2)^3)^2 )/ 2$ ?

1 Answer

Explanation:

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