Answer
$0.58$ hours
Work Step by Step
Plugging in $k=0.25,C=0.9C_0$ into the formula, we get:
$t=-0.25\ln(1-\frac{0.9C_0}{C_0})=-0.25\ln(1-0.9)\approx0.58$ hours
College Algebra 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305115546
ISBN 13:
978-1-30511-554-5
Chapter 4, Exponential and Logarithmic Functions - Section 4.3 - Logarithmic Functions - 4.3 Exercises - Page 389: 100
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