Answer
$15679$ years
Work Step by Step
$A=A_{0}e^{kt}$
For $A=1e^{-0.000121t}, $
$0.15=1e^{-0.000121t}$
$\ln0.15=-0.000121t $
$t\approx 15679$ 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: 19
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