Using laws of exponents, simplify and then evaluate: 1. (33⋅3)2+[(2)5(−2)2]3 2. (−3)7(−3)2(−3)3

1.&nbsp;(33⋅3)2+[(2)5(−2)2]3(33⋅3)2+[(2)5(−2)2]3=(33×31)2+[25(−2)2]3{33+1}2+{[25(−1)222]}3=(34)2+[2522]3=34×2+[25×2−2]3=38+[25−2]3=38+[23]3=38+23×3=38+29=6561+512=70732.&nbsp;(−3)7(−3)2(−3)3=(−1)7(3)7(−1)2(3)2(−1)3(3)3=−371×32×−1×33=−37−32×33=3732+3=3735=37×3−5=37−5=32=9Hence the value of&nbsp;(−3)7(−3)2(−3)3

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High School
Algebra
Algebra I
Exponents and radicals

Lennie Carroll

Answered question

2021-10-11

Using laws of exponents, simplify and then evaluate:
1.

{(3^{3} \cdot 3)}^{2} + {[\frac{{(2)}^{5}}{{(- 2)}^{2}}]}^{3}

\frac{{(- 3)}^{7}}{{(- 3)}^{2}} {(- 3)}^{3}

Answer & Explanation

Jayden-James Duffy

Skilled2021-10-12Added 91 answers

1. ${(3^{3} \cdot 3)}^{2} + {[\frac{{(2)}^{5}}{{(- 2)}^{2}}]}^{3}$
${(3^{3} \cdot 3)}^{2} + {[\frac{{(2)}^{5}}{{(- 2)}^{2}}]}^{3} = {(3^{3} \times 3^{1})}^{2} + {[\frac{2^{5}}{{(- 2)}^{2}}]}^{3}$
${3^{3 + 1}}^{2} + {[\frac{2^{5}}{(- 1)^{22^{2}}}]}^{3}$
$= {(3^{4})}^{2} + {[\frac{2^{5}}{2^{2}}]}^{3}$
$= 3^{4 \times 2} + {[2^{5} \times 2^{- 2}]}^{3}$
$= 3^{8} + {[2^{5 - 2}]}^{3}$
$= 3^{8} + {[2^{3}]}^{3}$
$= 3^{8} + 2^{3 \times 3}$
$= 3^{8} + 2^{9}$
$= 6561 + 512$
$= 7073$
2. $\frac{{(- 3)}^{7}}{{(- 3)}^{2} {(- 3)}^{3}} = \frac{{(- 1)}^{7} {(3)}^{7}}{{(- 1)}^{2} {(3)}^{2} {(- 1)}^{3} {(3)}^{3}}$
$= \frac{- 3^{7}}{1 \times 3^{2} \times - 1 \times 3^{3}}$
$= \frac{- 3^{7}}{- 3^{2} \times 3^{3}}$
$= \frac{3^{7}}{3^{2 + 3}}$
$= \frac{3^{7}}{3^{5}}$
$= 3^{7} \times 3^{- 5}$
$= 3^{7 - 5}$
$= 3^{2}$
$= 9$
Hence the value of $\frac{{(- 3)}^{7}}{{(- 3)}^{2} {(- 3)}^{3}}$

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