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: 70



Work Step by Step

We solve: $\log_{8}(x+5)-\log_{8}(x-2)=1$ $\log_{8}\frac{x+5}{x-2}=1$ $8^1=\frac{x+5}{x-2}$ $x+5=8(x-2)$ $x+5=8x-16$ $7x=21 $ $x=3$
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