Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 12 Sequences and Series - Cumulative Review - Page 849: 47


After approximately five years.

Work Step by Step

From the exercise, the car's value after $n$ years is: $18600(1-0.155)^n$, hence our equation is: $18600(1-0.155)^n=8000\\0.845^n=\frac{8000}{18600}=\frac{40}{93}\\n=\log_{0.845}{\frac{40}{93}}\approx5.00966$ Thus, the car will have that value after approximately five years.
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