College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.3: 37

Answer

$$\ln\Bigg[\frac{x^3\sqrt{x^2+1}}{(x+1)^4}\Bigg]=3\ln x+\frac{1}{2}\ln(x^2+1)-4\ln(x+1)$$

Work Step by Step

$$A=\ln\Bigg[\frac{x^3\sqrt{x^2+1}}{(x+1)^4}\Bigg]$$ First, we apply the Quotient Rule, which states $$\log_b\frac{M}{N}=\log_b M-\log_bN$$ ($M, N, b\in R, M\gt0, N\gt0, b\gt0, b\ne1$) That means, $$A=\ln(x^3\sqrt{x^2+1})-\ln(x+1)^4$$ Now, for $\ln(x^3\sqrt{x^2+1})$, we apply the Product Rule, which states $$\log_bMN=\log_bM+\log_bN$$ ($M, N, b\in R, M\gt0, N\gt0, b\gt0, b\ne1$) Therefore, $$A=\ln(x^3)+\ln \sqrt{x^2+1}-\ln(x+1)^4$$ $$A=\ln(x^3)+\ln(x^2+1)^{1/2}-\ln(x+1)^4$$ Finally, Power Rule can be applied for all, $$\log_bM^p=p\log_bM$$ ($M, b, p\in R, M\gt0, b\gt0, b\ne1$) So, $$A=3\ln x+\frac{1}{2}\ln(x^2+1)-4\ln(x+1)$$
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