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: 21


$x=-2$ and $x=1$

Work Step by Step

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