## Thomas' Calculus 13th Edition

a) $\approx 8$ years b) $\approx 32.02$ years
a) Consider the exponential growth as: $y=y_0e^{kt}$ As we are given that $y=1000$ and $y_0=10,000$ Then, we get $1,000=10,000 e^{(\ln 0.75)t}$ or, $t=\dfrac{\ln (0.1)}{\ln (0.75)}$ Thus, $t \approx 8$ years b) Consider the exponential growth as: $y=y_0e^{kt}$ As we are given that $y=1$ and $y_0=10,000$ Thus, we have $(1)=(10,000) e^{(\ln 0.75)t}$ or, $t=\dfrac{\ln (0.0001)}{\ln (0.75)}$ Thus, $t \approx 32.02$ years