Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 27

$t=\frac{\ln(2)}{0.06}\approx 11.55$

We have: $$P(t)=2.4e^{0.06t}$$ $$P(0)=2.4e^{0.06cdot 0}=2.4$$ Find $t$ such that: $$P(t)=2P(0)$$ $$2.4e^{0.06t}=2\cdot 2.4$$ $$e^{0.06t}=2$$ $$\ln(e^{0.06t})=\ln(2)$$ $$0.06t=\ln(2)$$ $$t=\frac{\ln(2)}{0.06}$$ $$t\approx 11.55$$