Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Section 4.4 - Laws of Logarithms - 4.4 exercises - Page 359: 70

Answer

$\dfrac{\log{7}}{\log{2}}$

Work Step by Step

RECALL: The change-of-base formula for loagarithms: $\log_b{x} = \dfrac{\log_a{x}}{\log_a{b}}$ Use the formula above to obtain: $(\log_2{5})(\log_5{7}) = \left(\dfrac{\log{5}}{\log{2}}\right)\left(\dfrac{\log{7}}{\log{5}}\right)$ Cancel common factors to obtain: $\require{cancel}=\left(\dfrac{\cancel{\log{5}}}{\log{2}}\right)\left(\dfrac{\log{7}}{\cancel{\log{5}}}\right) \\=\dfrac{\log{7}}{\log{2}}$
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