Prealgebra (7th Edition)

Published by Pearson
ISBN 10: 0321955048
ISBN 13: 978-0-32195-504-3

Chapter 10 - Section 10.2 - Multiplication Properties of Exponents - Exercise Set - Page 711: 49

Answer

(They ask you to write the answer in your own words.) You can get $x^{2}$ $\times$ $x^{3}$ = $x^{5}$ by using the exponent product rule . $3$ $+$ $2$ $=$ $5$. Why does this work, though? $x^{2}$ is $x$ $\times$ $x$. $x^{3}$ is $x$ $\times$ $x$ $\times$ $x$ So essentially, the problem $x^{2}$ $\times$ $x^{3}$ = $x^{5}$ is $x$ $\times$ $x$ $\times$ $x$ $\times$ $x$ $\times$ $x$. And that is $x^{5}$. How would you solve ($x^{2}$)$^{3}$? Using the exponent power rule, you would multiply $3$ $\times$ $2$ $=$ $6$. And you would get $x^{6}$. But why does this rule work? ($x^{2}$)$^{3}$ is basically $x^{2}$ $\times$ $x^{2}$ $\times$ $x^{2}$ So that is basically $x$ $\times$ $x$ $\times$ $x$ $\times$ $x$ $\times$ $x$ $\times$ $x$ And that equals $x^{6}$.

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