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: 19

Answer

$n=11$

Work Step by Step

$a_n=2(3)^n-24$ when $a_n=354270$ $2(3)^n-24=354270$ $\implies 2(3)^n=354294$ $\implies (3)^n=177147$ $\implies n=\log_3{177147}=11$
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