Answer
$7.084$ years
Work Step by Step
$A=A_{0}e^{kt}$
For $A=0.5, A_{0}=1, t=22 $ year
$0.5=1e^{22k} $
$\ln0.5=22k $
$k=-0.0315$ decay rate or $3.15\%$ decay per year.
Thus, $0.8=e^{-0.0315t}$
$\ln 0.8 =-0.0315t$
$t=7.084$ years
Intermediate Algebra for College Students (7th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-13417-894-7
ISBN 13:
978-0-13417-894-3
Chapter 9 - Section 9.6 - Exponential Growth and Decay; Modeling Data - Exercise Set - Page 741: 29
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