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

(a) $t=-10\ln 3$ (b) $t=\frac{1}{k}\ln\frac{1}{2}$ (c) $t=1-\log_{5}2$
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) $$e^{-0.3t}=27$$ - Take the natural logarithm of both sides: $$\ln e^{-0.3t}=\ln27$$ $$-0.3t=\ln27$$ $$t=\frac{\ln27}{-0.3}$$ - However, $\ln27=\ln3^3=3\ln3$ (Power Rule) Therefore, $$t=\frac{3\ln3}{-0.3}=\frac{\ln3}{-0.1}=-10\ln3$$ (b) $$e^{kt}=\frac{1}{2}$$ - Take the natural logarithm of both sides: $$\ln e^{kt}=\ln\frac{1}{2}$$ $$kt=\ln\frac{1}{2}$$ $$t=\frac{1}{k}\ln\frac{1}{2}$$ (c) $$e^{(\ln0.2)t}=0.4$$ - Take the natural logarithm of both sides: $$\ln e^{(\ln0.2)t}=\ln0.4$$ $$(\ln0.2)t=\ln0.4$$ $$t=\frac{\ln0.4}{\ln0.2}$$ - Recall the Change of Base Formula here: $$t=\log_{0.2}0.4=\log_{\frac{1}{5}}0.4=\log_{5^{-1}}0.4$$ - Here, we apply this property: $\log_{a^x}{b}=\frac{1}{x}\log_a{b}=\log_a{b^{\frac{1}{x}}}$ Therefore, $$t=\log_{5}(0.4)^{\frac{1}{-1}}=\log_{5}(0.4)^{-1}=\log_{5}\frac{1}{0.4}=\log_{5}\frac{5}{2}$$ - Apply Quotient Rule: $$t=\log_{5}5-\log_{5}2=1-\log_{5}2$$