Answer
a) $ -0.00001$ b) $10,536$ years c) $82$%
Work Step by Step
a) As we are given that $y=y_0e^{kt}$
Re-write as: $0.99 y_0=y_0e^{(1000)k}$
and $k=\dfrac{ \ln (0.99) }{1000} \implies k \approx -0.00001$
b) As we are given that $0.9=e^{-(0.00001)t}$
Re-write as: $t=\dfrac{ \ln (0.9) }{-0.00001}$
so, $ t \approx 10,536$ years
c) As we are given that $y=y_0e^{20,000k}$
Re-write as: $(y_0)e^{-0.2}=y_0(0.82)$
so, $y=82$%
