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

(a) $e^{\ln(x^2+y^2)}= x^2+y^2$ (b) $e^{-\ln 0.3}=\frac{10}{3}$ (c) $e^{\ln \pi x-\ln 2}=\frac{\pi x}{2}$
This whole exercise is based on this property: $$e^{\ln x}=x\hspace{1cm}x\gt0$$ (a) $$e^{\ln(x^2+y^2)}= x^2+y^2$$ (b) $$e^{-\ln 0.3}$$ - Apply Reciprocal Rule for $-\ln 0.3$: $-\ln 0.3=\ln\frac{1}{0.3}=\ln\frac{10}{3}$ $$e^{-\ln 0.3}=e^{\ln\frac{10}{3}}=\frac{10}{3}$$ (c) $$e^{\ln \pi x-\ln 2}$$ - Apply Quotient Rule here: $\ln \pi x-\ln 2 =\ln\frac{\pi x}{2}$ $$e^{\ln \pi x-\ln 2}=e^{\ln\frac{\pi x}{2}}=\frac{\pi x}{2}$$