Answer
(a) $\ln\sin\theta-\ln\frac{\sin\theta}{5}=\ln5$
(b) $\ln(3x^2-9x)+\ln\frac{1}{3x}=\ln(x-3)$
(c) $\frac{1}{2}\ln(4t^4)-\ln b=\ln\frac{2t^2}{b}$
Work Step by Step
(a) $$\ln\sin\theta-\ln\frac{\sin\theta}{5}$$
- Apply Quotient Rule: $$\ln\sin\theta-\ln\frac{\sin\theta}{5}=\ln\sin\theta-(\ln\sin\theta-\ln5)=\ln\sin\theta-\ln\sin\theta+\ln5$$ $$\ln\sin\theta-\ln\frac{\sin\theta}{5}=\ln5$$
(b) $$\ln(3x^2-9x)+\ln\frac{1}{3x}=\ln[3x(x-3)]+\ln\frac{1}{3x}$$
- Apply Product Rule for $\ln[3x(x-3)]$ and Reciprocal Rule for $\ln\frac{1}{3x}$:
$$\ln(3x^2-9x)+\ln\frac{1}{3x}=\ln3x+\ln(x-3)-\ln3x$$ $$\ln(3x^2-9x)+\ln\frac{1}{3x}=\ln(x-3)$$
(c) $$\frac{1}{2}\ln(4t^4)-\ln b$$
- Apply Power Rule: $\frac{1}{2}\ln(4t^4)=\ln(4t^4)^{1/2}=\ln2t^2$
Therefore, $$\frac{1}{2}\ln(4t^4)-\ln b=\ln2t^2-\ln b$$
- Apply Quotient Rule here: $$\frac{1}{2}\ln(4t^4)-\ln b=\ln\frac{2t^2}{b}$$
