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 23: 60

Answer

$a)$ $9$ $b)$ $\dfrac{1}{7}$ $c)$ $\dfrac{1}{36}$

Work Step by Step

$a)$ $3^{2/7}\cdot3^{12/7}$ Evaluate the product of powers and simplify: $3^{2/7}\cdot3^{12/7}=3^{2/7+12/7}=3^{14/7}=3^{2}=9$ $b)$ $\dfrac{7^{2/3}}{7^{5/3}}$ Evaluate the division and simplify: $\dfrac{7^{2/3}}{7^{5/3}}=7^{2/3-5/3}=7^{-3/3}=7^{-1}=\dfrac{1}{7}$ $c)$ $(\sqrt[5]{6})^{-10}$ Write this expression as a fraction to change the sign of the exponent: $(\sqrt[5]{6})^{-10}=\dfrac{1}{(\sqrt[5]{6})^{10}}=...$ Rewrite the radical expression as a power with rational exponent and simplify: $...=\dfrac{1}{(6^{1/5})^{10}}=\dfrac{1}{6^{10/5}}=\dfrac{1}{6^{2}}=\dfrac{1}{36}$
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