How do you simplify ${(2^{2})}^{2} \cdot {(2^{3})}^{- 1}$ $(2^2)^2*(2^3)^-1$ and write it using only positive exponents?

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Answer 1 · 2017-04-11T16:39:14

I found $2$ $2$

We can use some properties of exponents and write it as:
$\frac{2^{2 \cdot 2}}{2^{3}} = \frac{2^{4}}{2^{3}} = 2$ $2^(2*2)/2^3=2^4/2^3=2$

Where we used the following:
${(x^{a})}^{b} = x^{a \cdot b}$ $(x^a)^b=x^(a*b)$
and
$x^{- a} = \frac{1}{x^{a}}$ $x^-a=1/x^a$

Answer 2 · 2017-04-11T16:39:32

$2$ $2$

${(2^{2})}^{2} \cdot {(2^{3})}^{- 1}$ $(2^2)^2*(2^3)^-1$

$:.=(2*2)*(2*2)*(2*2*2)^(-1)$

$:.=4*4*8^-1$

$:.=16*1/8$

$:.=16/8$

$:.=2$

How do you simplify (2^2)^2*(2^3)^-1(22)2⋅(23)−1(2^2)^2*(2^3)^-1 and write it using only positive exponents?