College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Section 6.3 - Exponential Functions - 6.3 Assess Your Understanding - Page 436: 100

Answer

a) $F(-5)=240$ The point is (-5, 240). b) $x=-3$ The point is (-3, 24). c) $x=-1$

Work Step by Step

a) $F(-5)=(\frac{1}{3})^{-5}-3=3^5-3=243-3=240$ The point is (-5, 240). b) $24=(\frac{1}{3})^{x}-3$ $27=(\frac{1}{3})^{x}$ $(\frac{1}{27})^{-1}=(\frac{1}{3})^{x}$ $(\frac{1}{3})^{-3}=(\frac{1}{3})^{x}$ $x=-3$ The point is (-3, 24). c) $0=(\frac{1}{3})^{x}-3$ $3=(\frac{1}{3})^{x}$ $(\frac{1}{3})^{-1}=(\frac{1}{3})^{x}$ $x=-1$
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