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

(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) $\frac{\ln35+\ln\frac{1}{7}}{\ln25}=\frac{1}{2}$
(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: $$\frac{\ln35+\ln\frac{1}{7}}{\ln25}=\frac{\ln(5\times7)+\ln\frac{1}{7}}{\ln5^2}=\frac{\ln5+\ln7-\ln7}{2\ln5}=\frac{\ln5}{2\ln5}=\frac{1}{2}$$