Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 4 - Review - Exercises - Page 389: 70



Work Step by Step

$\log_{8}(x+5)-\log_{8}(x-2)=1$ Combine the subtraction on the left side as the logarithm of a division: $\log_{8}\dfrac{x+5}{x-2}=1$ Rewrite this equation in exponential form: $\dfrac{x+5}{x-2}=8^{1}$ $\dfrac{x+5}{x-2}=8$ Solve for $x$: $x+5=8(x-2)$ $x+5=8x-16$ $x-8x=-16-5$ $-7x=-21$ $x=\dfrac{-21}{-7}$ $x=3$
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