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 - Page 425: 42

Answer

$3$

Work Step by Step

Use the rule "$\log_a{P} - \log_a{Q} = \log_a{\left(\frac{P}{Q}\right)}$" to obtain: $=\log_5{\left(\frac{250}{2}\right)} \\=\log_5{\left(125\right)}$ Note that $125=5^3$. Thus, the expression above is equivalent to: $=\log_5{(5^3)}$ RECALL: $\log_a{(a^x)} = x$ Use the rule above to obtain: $\log_5{(5^3)} = 3$
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