Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 12 - Section 12.8 - Exponential and Logarithmic Equations and Problem Solving - Exercise Set - Page 896: 28


$x=-2$ and $x=4$

Work Step by Step

$\log_{8}(x^{2}-2x)=1$ Rewrite in exponential form: $8^{1}=x^{2}-2x$ $x^{2}-2x=8$ Take the $8$ to the left side of the equation": $x^{2}-2x-8=0$ Solve this equation by factoring: $(x+2)(x-4)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x+2=0$ $x=-2$ $x-4=0$ $x=4$ The initial equation is not undefined for either of the solutions found. The answer is $x=-2$ and $x=4$
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