Answer
(a) $t=-10\ln 3$
(b) $t=\frac{1}{k}\ln\frac{1}{2}$
(c) $t=1-\log_{5}2$
Work Step by Step
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: $$t=\log_{5}5-\log_{5}2=1-\log_{5}2$$
