## University Calculus: Early Transcendentals (3rd Edition)

(a) $t=-300\ln10$ (b) $t=-\frac{\ln10}{k}$ (c) $t=-1$
Another method to solve these exercises is to take the natural logarithm of both sides. In other words: $$e^x=a\hspace{1cm}\text{then}\hspace{1cm}\ln e^{x}=\ln a\hspace{1cm}\text{then}\hspace{1cm} x=\ln a$$ (a) $$e^{-0.01t}=1000$$ - Take the natural logarithm of both sides: $$\ln e^{-0.01t}=\ln1000$$ $$-0.01t=\ln1000$$ $$t=\frac{\ln1000}{-0.01}$$ - However, $\ln27=\ln10^3=3\ln10$ (Power Rule) Therefore, $$t=\frac{3\ln10}{-0.01}=-300\ln10$$ (b) $$e^{kt}=\frac{1}{10}$$ - Take the natural logarithm of both sides: $$\ln e^{kt}=\ln\frac{1}{10}$$ $$kt=\ln\frac{1}{10}$$ $$t=\frac{1}{k}\ln\frac{1}{10}$$ - Apply Reciprocal Rule: $$t=\frac{1}{k}(-\ln10)=-\frac{\ln10}{k}$$ (c) $$e^{(\ln2)t}=\frac{1}{2}$$ - Take the natural logarithm of both sides: $$\ln e^{(\ln2)t}=\ln\frac{1}{2}$$ $$(\ln2)t=\ln\frac{1}{2}$$ $$t=\frac{\ln\frac{1}{2}}{\ln2}$$ - Apply Reciprocal Rule here: $$t=\frac{-\ln2}{\ln2}=-1$$