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

(a) $\ln0.75=\ln3-2\ln2$ (b) $\ln\frac{4}{9}=2\ln2-2\ln3$ (c) $\ln\frac{1}{2}=-\ln2$ (d) $\ln\sqrt[3]9=\frac{2}{3}\ln3$ (e) $\ln3\sqrt2=\ln3+\frac{1}{2}\ln2$ (f) $\ln\sqrt{13.5}=\frac{3}{2}\ln3-\frac{1}{2}\ln2$
(a) $$\ln0.75= \ln\Big(\frac{3}{4}\Big)$$ - Apply Quotient Rule: $$\ln0.75=\ln3-\ln4=\ln3-\ln2^2$$ - Apply Power Rule: $$\ln0.75=\ln3-2\ln2$$ (b) $$\ln\frac{4}{9}$$ - Apply Quotient Rule: $$\ln\frac{4}{9}=\ln4-\ln9=\ln2^2-\ln3^2$$ - Apply Power Rule: $$\ln\frac{4}{9}=2\ln2-2\ln3$$ (c) $$\ln\frac{1}{2}$$ - Apply Reciprocal Rule: $$\ln\frac{1}{2}=-\ln2$$ (d) $$\ln\sqrt[3]9=\ln9^{1/3}=\ln(3^2)^{1/3}=\ln3^{2/3}$$ - Apply Power Rule: $$\ln\sqrt[3]9=\frac{2}{3}\ln3$$ (e) $$\ln3\sqrt2$$ - Apply Product Rule: $$\ln3\sqrt2=\ln3+\ln\sqrt2=\ln3+\ln2^{1/2}$$ - Apply Power Rule: $$\ln3\sqrt2=\ln3+\frac{1}{2}\ln2$$ (f) $$\ln\sqrt{13.5}=\ln(13.5)^{1/2}$$ - Apply Power Rule: $$\ln\sqrt{13.5}=\frac{1}{2}\ln13.5=\frac{1}{2}\ln\sqrt\frac{27}{2}$$ - Apply Quotient Rule: $$\ln\sqrt{13.5}=\frac{1}{2}(\ln27-\ln2)=\frac{1}{2}(\ln3^3-\ln2)$$ - Apply Power Rule: $$\ln\sqrt{13.5}=\frac{1}{2}(3\ln3-\ln2)=\frac{3}{2}\ln3-\frac{1}{2}\ln2$$