Answer
(a) $\ln\frac{1}{125}=-3\ln5$
(b) $\ln9.8=2\ln7-\ln5$
(c) $\ln7\sqrt7=\frac{3}{2}\ln7$
(d) $\ln1225=2\ln5+2\ln7$
(e) $\ln0.056=\ln7-3\ln5$
(f) $\frac{\ln35+\ln\frac{1}{7}}{\ln25}=\frac{1}{2}$
Work Step by Step
(a) $$\ln\frac{1}{125}$$
- Apply Reciprocal Rule: $$\ln\frac{1}{125}=-\ln125=-\ln5^3$$
- Apply Power Rule: $$\ln\frac{1}{125}=-3\ln5$$
(b) $$\ln9.8=\ln\frac{49}{5}$$
- Apply Quotient Rule: $$\ln9.8=\ln49-\ln5=\ln7^2-\ln5$$
- Apply Power Rule: $$\ln9.8=2\ln7-\ln5$$
(c) $$\ln7\sqrt7=\ln(7\times7^{1/2})=\ln7^{3/2}$$
- Apply Power Rule: $$\ln7\sqrt7=\frac{3}{2}\ln7$$
(d) $$\ln1225=\ln(25\times49)$$
- Apply Product Rule: $$\ln1225=\ln25+\ln49=\ln5^2+\ln7^2$$
- Apply Power Rule: $$\ln1225=2\ln5+2\ln7$$
(e) $$\ln0.056=\ln\Big(\frac{7}{125}\Big)$$
- Apply Quotient Rule: $$\ln0.056=\ln7-\ln125=\ln7-\ln5^3$$
- Apply Power Rule: $$\ln0.056=\ln7-3\ln5$$
(f) $$\frac{\ln35+\ln\frac{1}{7}}{\ln25}=\frac{\ln(5\times7)+\ln\frac{1}{7}}{\ln5^2}$$
- For $\ln(5\times7)$, apply Product Rule: $\ln(5\times7)=\ln5+\ln7$
- For $\ln\frac{1}{7}$, apply Reciprocal Rule: $\ln\frac{1}{7}=-\ln7$
- For $\ln5^2$, apply Power Rule: $\ln5^2=2\ln5$
Therefore: $$\frac{\ln35+\ln\frac{1}{7}}{\ln25}=\frac{\ln(5\times7)+\ln\frac{1}{7}}{\ln5^2}=\frac{\ln5+\ln7-\ln7}{2\ln5}=\frac{\ln5}{2\ln5}=\frac{1}{2}$$
