Answer
See below.
Work Step by Step
We know that $x=2xe^{k\cdot1690}\\0.5=e^{5730k}\\5730k=\ln0.5\\k=\frac{\ln0.5}{5730}\approx-0.000120968$
Thus $0.05=e^{-0.000120968\cdot t}\\-0.000120968t=\ln0.05\\t=\frac{\ln0.05}{-0.000120968}\approx24765$
College Algebra (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0321979478
ISBN 13:
978-0-32197-947-6
Chapter 6 - Review Exercises - Page 500: 51
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