Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 12 - Review - Exercises - Page 890: 31

Answer

$12,288$

Work Step by Step

Observing the number of bacteria after 0s, 5s, 10s, 15s,... we define a sequence with $a_{n}$ = the number of bacteria after $5(n-1)$ seconds. $a_{1}=a=3$, $a_{2}=3\cdot 2=a\cdot 2$, $a_{3}=3\cdot 2^{2}=a\cdot 2^{2}$, $a_{4}=3\cdot 2^{3}=a\cdot 2^{3}$, $\ldots$. We obtain a geometric sequence with $r=2$ and $a=3$. $a_{n}=3\cdot 2^{n-1}$ So after $60=5(13-1)$ seconds, the number of bacteria is $a_{13}=3\cdot 2^{12}=12,288$.
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