Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 9 - Conic Sections, Sequences, and Series - 9.3 Arithmetic Sequences - 9.3 Exercises - Page 727: 20

Answer

$n=12$

Work Step by Step

$a_n=2048\left(\frac{1}{2}\right)^n-80$ when $a_n=-79.5$ $2048\left(\frac{1}{2}\right)^n-80=-79.5$ $\implies 2048\left(\frac{1}{2}\right)^n=0.5$ $\implies\left(\frac{1}{2}\right)^n=\frac{1}{4096}$ $\implies n=\log_{\frac{1}{2}}\frac{1}{4096}=12$
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