Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Cumulative Review Exercises - Page 709: 2


Using the logarithm, we have $x-1=log_{11}(125)=\frac{log125}{log11}\approx2.0136$; thus $x=1+\frac{log125}{log11}\approx3.01$

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