Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.2 - Exponents and Radicals - 1.2 Exercises - Page 21: 8

Answer

a) No b) No c) Yes d) No

Work Step by Step

a) ($x^{2}$)$^{3}$ = $x^{2\times3}$ = $x^{6}$ $\ne$ $x^{5}$ b) ($2x^{4}$)$^{3}$ = $2^{3}$ $\times$ $x^{4\times3}$ = 8$x^{12}$ c) $\sqrt (4a^{2}$) = $\sqrt 4$ $\times$ $\sqrt a^{2}$ = 2a d) One way to see that these two numbers are different is by comparing their squares: ($\sqrt (a^{2}+4$))$^{2}$ = $a^{2}+4$ $\ne$ (a+2)$^{2}$ = a$^{2}$ + 4 + 4a
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