University Calculus: Early Transcendentals (3rd Edition)

a) $-0.00001$ b) $10,536$ years c) $82$%
a) Given: $y=y_0e^{kt}$ Re-arrange the given equation as follows:. Then, $0.99 y_0=y_0e^{1000k}$ Then $k=\dfrac{ \ln (0.99) }{1000} \approx -0.00001$ b) Given: $0.9=e^{-0.00001t}$ Re-arrange the given equation as follows: Then $t=\dfrac{ \ln (0.9) }{-0.00001} \approx 10,536$ years c) Given: $y=y_0e^{20,000k}$ Re-arrange the given equation as follows:. Then, $y_0e^{-0.2}=y_0(0.82)$ Thus, $y=82$% Hence, a) $-0.00001$ b) $10,536$ years c) $82$%