Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 5 - Exponential and Logarithmic Functions - 5.5 Properties of Logarithms - 5.5 Assess Your Understanding - Page 306: 97

Answer

$3$

Work Step by Step

We know that $\log_a {b}=\frac{\log_c {b}}{\log_c {a}}$ (this is known as the Change-of-Base formula). Hence, $ \log_2 {3}\cdot \log_3 {4}\cdot \log_4 {5}\cdot \log_5 {6}\cdot \log_6 {7}\cdot \log_7 {8}\\$ Cancel the common factors to obtain $=\dfrac{\log{3}}{\log{2}}\cdot\dfrac{\log{4}}{\log{3}}\cdot\dfrac{\log{5}}{\log{4}}\cdot\dfrac{\log{6}}{\log{5}}\cdot\dfrac{\log{7}}{\log{6}}\cdot\dfrac{\log{8}}{\log{7}}\\ =\dfrac{\log{8}}{\log{2}}\\$ Use the Change-of-Base formula to obtain $\dfrac{\log{8}}{\log{2}}=\log_2 {8}=\log_2 {2^3}$ We know that $\log_a {x^n}=n\cdot \log_a {x}$. Hence, $\log_2 {2^3}=3\log_2 {2}=3\cdot1=3$
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