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

Published by Pearson

# Chapter 1 - Section 1.6 - Inverse Functions and Logarithms - Exercises - Page 49: 43

#### 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: $$\frac{1}{2}\ln(4t^4)-\ln b=\ln\frac{2t^2}{b}$$

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