Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.3 Logarithmic Functions - 4.3 Exercises - Page 446: 74


$1+\dfrac {1}{2}\log _{2}3-\log _{2}5$

Work Step by Step

$$\log _{2}\dfrac {2\sqrt {3}}{5}=\log _{2}2^{1}\times 3^{\dfrac {1}{2}}\times 5^{-1}=\log _{2}2+\log _{2}3^{1/2}-\log _{2}5=1+\dfrac {1}{2}\log _{2}3-\log _{2}5$$
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