College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 4, Exponential and Logarithmic Functions - Chapter 4 Review - Concept Check: 39

Answer

$2$

Work Step by Step

RECALL: (1) $\log{P} + \log{Q} = \log{(PQ)}, P, Q \gt 0$ (2) $\log{(10^k)} = k$ Use rule (1) above to obtain: $=\log{(25\cdot4)} \\=\log{100}$ Note that $100=10^2$. Thus, the expression above is equivalent to: $\log{100} = \log{(10^2)}$ Use rule (2) above to obtain: $\log{(10^2)} = 2$
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