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: 65



Work Step by Step

$$A=\frac{1}{2}(\log_5x+\log_5y)-2\log_5(x+1)$$ First, we know from the Product Rule that $$\log_bM+\log_bN=\log_bMN$$ ($M, N, b\in R, M\gt0, N\gt0, b\gt0, b\ne1$) Apply it to $\log_5x+\log_5y$, we have $$A=\frac{1}{2}\log_5(xy)-2\log_5(x+1)$$ Now, from the Power Rule that $$p\log_bM=\log_bM^p$$ ($M, b, p\in R, M\gt0, b\gt0, b\ne1$), we also can deduce $$A=\log_5(xy)^{1/2}-\log_5(x+1)^2$$ $$A=\log_5\sqrt{xy}-\log_5(x+1)^2$$ Finally, we know the Quotient Rule, which states $$\log_b M-\log_bN=\log_b\frac{M}{N}$$ ($M, N, b\in R, M\gt0, N\gt0, b\gt0, b\ne1$) That means, $$A=\log_5\frac{\sqrt{xy}}{(x+1)^2}$$
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