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

(a) $k=\frac{1}{5}\ln\frac{1}{4}$ (b) $k=\ln\frac{1}{80}$ (c) $k=1$
To solve for $k$ in this exercise, keep in mind this property: - If $e^x = e^a$ then $x=a$ (a) $$e^{5k}=\frac{1}{4}$$ - Recall this inverse property: $e^{\ln x}=x$ Therefore, we can rewrite $\frac{1}{4}=e^{\ln\frac{1}{4}}$ That means $$e^{5k}=e^{\ln\frac{1}{4}}$$ $$5k=\ln\frac{1}{4}$$ $$k=\frac{1}{5}\ln\frac{1}{4}$$ (b) $$80e^{k}=1$$ $$e^{k}=\frac{1}{80}$$ - Recall this inverse property: $e^{\ln x}=x$ Therefore, we can rewrite $\frac{1}{80}=e^{\ln\frac{1}{80}}$ That means $$e^{k}=e^{\ln\frac{1}{80}}$$ $$k=\ln\frac{1}{80}$$ (c) $$e^{(\ln0.8)k}=0.8$$ - Recall this inverse property: $e^{\ln x}=x$ Therefore, we can rewrite $0.8=e^{\ln 0.8}$ That means $$e^{(\ln0.8)k}=e^{\ln 0.8}$$ $$(\ln0.8)k=\ln 0.8$$ $$k=1$$