Answer
(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) $\ln\sqrt{13.5}=\frac{3}{2}\ln3-\frac{1}{2}\ln2$
Work Step by Step
(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: $$\ln\sqrt{13.5}=\frac{1}{2}(3\ln3-\ln2)=\frac{3}{2}\ln3-\frac{1}{2}\ln2$$
