Answer
(a) $e^{\ln(x^2+y^2)}= x^2+y^2$
(b) $e^{-\ln 0.3}=\frac{10}{3}$
(c) $e^{\ln \pi x-\ln 2}=\frac{\pi x}{2}$
Work Step by Step
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: $\ln \pi x-\ln 2 =\ln\frac{\pi x}{2}$
$$e^{\ln \pi x-\ln 2}=e^{\ln\frac{\pi x}{2}}=\frac{\pi x}{2}$$
