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

(a) $2\ln\sqrt e=1$ (b) $\ln(\ln e^e)=1$ (c) $\ln e^{-x^2-y^2}=-x^2-y^2$
This whole exercise is based on this property: $$\ln e^x=x\hspace{1cm}x\gt0$$ (a) $$2\ln\sqrt e$$ - First, apply Power Rule: $$2\ln\sqrt e=\ln(\sqrt e)^2=\ln e=1$$ Therefore, $$2\ln\sqrt e=1$$ (b) $$\ln(\ln e^e)$$ - Considering $\ln e^e$ first: $$\ln e^e=e$$ Therefore, $$\ln(\ln e^e)=\ln e=1$$ So, $$\ln(\ln e^e)=1$$ (c) $$\ln e^{-x^2-y^2}=-x^2-y^2$$