Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 3 - Test - Page 516: 28


$ t \approx 12.5$ days

Work Step by Step

The exponential decay process can be written as: $ N(t) N_0 e^{-\lambda t}$ ...(1) But $\lambda =\dfrac{\ln 2}{t_{1/2}}= 0.096$ We need to simply equation (1) to obtain an expression for $ t $ when $ N(t)=0.3 N_0$ Plug in the given data: $0.3 N_0=N_0 e^{-0.096} t $ $\ln 0.3 =\ln (e^{-0.096t})$ or, $\ln 0.3 =-0.096t $ Thus, $ t \approx 12.5$ days
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