Chapter 7 - Section 7.2 - Exponential Change and Separable Differential Equations - Exercises - Page 411: 37

$\approx 56563$ years

Given: $A_0=10$ and $A=5$ The exponential growth can be written as: $A=A_0e^{kt}$ ...(1) This implies that $5=10e^{24360k} \implies k =\dfrac{\ln 0.5}{24360}$ or, $k\approx -0.000028454$ years Equation (1) becomes: $(0.2)(10)=10 e^{-0.000028454} t$ $\implies \dfrac{\ln (0.2)}{-0.000028454}$ or, $t\approx 56563$ years