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

(a) $\ln(e^{\sec\theta})=\sec\theta$ (b) $\ln e^{e^x}=e^x$ (c) $\ln (e^{2\ln x})=2\ln x$
This whole exercise is based on this property: $$\ln e^x=x\hspace{1cm}x\gt0$$ (a) $$\ln(e^{\sec\theta})=\ln e^{\sec\theta}=\sec\theta$$ Therefore, $$\ln(e^{\sec\theta})=\sec\theta$$ (b) $$\ln e^{e^x}=\ln e^{(e^x)}=e^x$$ Therefore, $$\ln e^{e^x}=e^x$$ (c) $$\ln (e^{2\ln x})=\ln e^{2\ln x}=2\ln x$$ Therefore, $$\ln (e^{2\ln x})=2\ln x$$