Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 4 - Section 4.3 - Logarithmic Functions - 4.3 Exercises - Page 352: 30

Answer

a.) 3 b.) $\frac{1}{2}$ c.) $\frac{1}{4}$

Work Step by Step

$Evaluate$ $the$ $expression:$ a.) $\log_5 125$ b.) $\log_{49} 7$ c.) $\log_9 \sqrt 3$ a.) $\log_5 125$ Rewrite 125 as $5^3$ [Note: $5^3 = 5\times5\times5 = 125$] $\log_5 5^3$ Use the Third Property of Logarithms: $\log_a a^x = x$ $\log_5 5^3 = 3$ b.) $\log_{49} 7$ Rewrite 7 as $49^{\frac{1}{2}}$ [Note: $49^{\frac{1}{2}} = \sqrt{49} = 7$] $\log_{49} 49^{\frac{1}{2}}$ Use the Third Property of Logarithms: $\log_a a^x = x$ $\log_{49} 49^{\frac{1}{2}} = \frac{1}{2}$ c.) $\log_9 \sqrt 3$ Rewrite 3 as $9^{\frac{1}{2}}$ [Note: $9^{\frac{1}{2}} = \sqrt 9 = 3$] $\log_9 \sqrt{9^{\frac{1}{2}}}$ Rewrite the root to exponential form $\log_9 (9^{\frac{1}{2}})^{\frac{1}{2}} \rightarrow \log_9 9^{\frac{1}{2}\times\frac{1}{2}}$ $\log_9 9^{\frac{1}{4}}$ Use the Third Property of Logarithms: $\log_a a^x = x$ $\log_9 9^{\frac{1}{4}} = \frac{1}{4}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.