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: 22


$x=3$ and $x=-2$

Work Step by Step

$\log_{6}(x^{2}-x)=1$ Rewrite in exponential form: $6^{1}=x^{2}-x$ $x^{2}-x=6$ Take the $6$ to the left side of the equation: $x^{2}-x-6=0$ Solve this equation by factoring: $(x-3)(x+2)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x-3=0$ $x=3$ $x+2=0$ $x=-2$ The initial equation is not undefined for either of the solutions, so our answer is: $x=3$ and $x=-2$
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