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: 67

Answer

$\ln\sqrt[3]{\frac{(x+5)^2}{x(x^2-4)}}$

Work Step by Step

$$A=\frac{1}{3}[2\ln(x+5)-\ln x-\ln(x^2-4)]$$ First, we need to make coefficients of all logarithms change into 1 before being able to apply the Quotient Rule. To do that, we use the Power Rule. We know from the Power Rule that $$p\log_bM=\log_bM^p$$ ($M, b, p\in R, M\gt0, b\gt0, b\ne1$) Apply it to $2\ln(x+5)$, we have $$A=\frac{1}{3}[\ln(x+5)^2-\ln x-\ln(x^2-4)]$$ $$A=\frac{1}{3}[\ln(x+5)^2-[\ln x+\ln(x^2-4)]]$$ Now we can apply the Product Rule for $[\ln x+\ln(x^2-4)]$, that states $$\log_bM+\log_bN=\log_bMN$$ ($M, N, b\in R, M\gt0, N\gt0, b\gt0, b\ne1$) $$A=\frac{1}{3}[\ln(x+5)^2-\ln x(x^2-4)]$$ From the Quotient Rule $$\log_b M-\log_bN=\log_b\frac{M}{N}$$ ($M, N, b\in R, M\gt0, N\gt0, b\gt0, b\ne1$), we can apply to A $$A=\frac{1}{3}\ln\Bigg[\frac{(x+5)^2}{x(x^2-4)}\Bigg]$$ Again, use the Power Rule here. $$A=\ln\Bigg[\frac{(x+5)^2}{x(x^2-4)}\Bigg]^{1/3}$$ $$A=\ln\sqrt[3]{\frac{(x+5)^2}{x(x^2-4)}}$$
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