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: 100



Work Step by Step

We know that $\log_a {x^n}=n\cdot \log_a {x}$ and $\log_a{a}=1$ Hence, $\log_2 {2}\cdot \log_2 {4}\cdot... \log_2 {2^n}\\ =\log_2{2} \cdot \log_2{2^2} \cdot \log_2{2^3} \cdot \log_2{2^4}...\log_2{2^n}\\ =(1\cdot \log_2 {2})\cdot(2\cdot \log_2 {2})\cdot...(n\cdot \log_2 {2})\\ =(1\cdot1) \cdot (2\cdot 1) \cdot (3\cdot 1)...(n\cdot n)\\ =1\cdot2 \cdot 3...\cdot n\\ =n!$
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