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 - Page 477: 40


$$\log\Bigg[\frac{100x^3\sqrt[3]{5-x}}{3(x+7)^2}\Bigg]=3\log x+\frac{1}{3}\log(5-x)-2\log(x+7)+2-\log3$$

Work Step by Step

$$A=\log\Bigg[\frac{100x^3\sqrt[3]{5-x}}{3(x+7)^2}\Bigg]$$ 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, $$A=\log(100x^3\sqrt[3]{5-x})-\log[3(x+7)^2]$$ Now 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, $$A=\log100+\log x^3+\log\sqrt[3]{5-x}-[\log3+\log(x+7)^2]$$ $$A=\log100+\log x^3+\log({5-x})^{1/3}-\log3-\log(x+7)^2$$ Finally, Power Rule can be applied for all, $$\log_bM^p=p\log_bM$$ ($M, b, p\in R, M\gt0, b\gt0, b\ne1$) Also, $\log M$ is in fact the short form of $\log_{10}M$. That makes $\log100=\log_{10}100=2$ since $10^2=100$. So, $$A=2+3\log x+\frac{1}{3}\log(5-x)-\log3-2\log(x+7)$$ $$A=3\log x+\frac{1}{3}\log(5-x)-2\log(x+7)+2-\log3$$
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