Answer
(a) $\ln \sec\theta+\ln\cos\theta=0$
(b) $\ln(8x+4)-2\ln c=\ln\frac{8x+4}{c^2}$
(c) $3\ln\sqrt[3]{t^2-1}-\ln(t+1)=\ln(t-1)$
Work Step by Step
(a) $$\ln \sec\theta+\ln\cos\theta$$
- Apply Product Rule: $$\ln \sec\theta+\ln\cos\theta=\ln(\sec\theta\times\cos\theta)=\ln\Big(\frac{1}{\cos\theta}\times\cos\theta\Big)=\ln1=0$$
So, $$\ln \sec\theta+\ln\cos\theta=0$$
(b) $$\ln(8x+4)-2\ln c$$
- Apply Power Rule: $$\ln(8x+4)-2\ln c=\ln(8x+4)-\ln c^2$$
- Apply Quotient Rule: $$\ln(8x+4)-2\ln c=\ln\frac{8x+4}{c^2}$$
(c) $$3\ln\sqrt[3]{t^2-1}-\ln(t+1)$$
- First, apply Power Rule: $$3\ln\sqrt[3]{t^2-1}-\ln(t+1)=\ln(\sqrt[3]{t^2-1})^3-\ln(t+1)$$ $$3\ln\sqrt[3]{t^2-1}-\ln(t+1)=\ln(t^2-1)-\ln(t+1)$$
- Here, apply Quotient Rule: $$3\ln\sqrt[3]{t^2-1}-\ln(t+1)=\ln\frac{t^2-1}{t+1}$$ $$3\ln\sqrt[3]{t^2-1}-\ln(t+1)=\ln\frac{(t-1)(t+1)}{t+1}$$ $$3\ln\sqrt[3]{t^2-1}-\ln(t+1)=\ln(t-1)$$
