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: 40

Answer

$\frac{5}{2}$

Work Step by Step

RECALL: (1) $\log_a{(a^k)} = k$ (2) $\sqrt{a} = a^{\frac{1}{2}}$ (3) $(a^m)^n=a^{mn}$ Use rule (2) above to obtain: $\log_3{\sqrt{243}}= \log_3{(243^{\frac{1}{2}})}$ Note that $243=3^5$. Thus, the expression above is equivalent to: $\log_3{(243^{\frac{1}{2}})} = \log_3{[(3^5)^{\frac{1}{2}}]}$ Use rule (3) above to obtain: $\log_3{[(3^5)^{\frac{1}{2}}]} \\=\log_3{(3^{\frac{5}{2}\cdot 5})} \\=\log_3{(3^{\frac{5}{2}})}$ Use rule (1) above to obtain: $\log_3{(3^{\frac{5}{2}})}=\frac{5}{2}$
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