Chapter 7: Transcendental Functions - Section 7.4 - Exponential Change and Separable Differential Equations - Exercises 7.4 - Page 402: 37

$\approx 56563$ years

Consider the exponential growth equation as follows: $A=A_0e^{kt}$ . As we are given that $A_0=10$ and $A=5$ This implies that $5=10e^{24360k}$ or, $k =\dfrac{\ln 0.5}{24360} \approx -0.000028454$ years Now, we have $A=A_0e^{kt} \implies (0.2)(10)=(10) e^{-(0.000028454)} t$ $ \dfrac{\ln (0.2)}{-0.000028454}$ Thus, we get $t\approx 56563$ years