Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 11 - Systems of Equations and Inequalities - Chapter Review - Cumulative Review - Page 797: 5



Work Step by Step

We are given the equation: $\log_3 (x-1)+\log_3 (2x+1)=2$ Use the Product Property of logarithms: $\log_3 (x-1)(2x+1)=2$ Rewrite the equation in exponential form: $(x-1)(2x+1)=3^2$ $2x^2+x-2x-1-9=0$ $2x^2-x-10=0$ Solve for $x$: $x=\dfrac{1\pm\sqrt{(-1)^2-4(2)(-10)}}{2(2)}=\dfrac{1\pm 9}{4}$ $x_1=\dfrac{1-9}{4}=-2$ $x_2=\dfrac{1+9}{4}=\dfrac{5}{2}$ Check the solutions: $x_1=-2$ $\log_3 (-2-1)+\log_3 (2(-2)+1)\stackrel{?}{=}2$ The logarithms are not defined for $x=-2$. $x_2=\dfrac{5}{2}$ $\log_3 \left(\dfrac{5}{2}-1\right)+\log_3 \left(2\cdot\dfrac{5}{2}+1\right)\stackrel{?}{=}2$ $\log_3 \dfrac{3}{2}+\log_3 6\stackrel{?}{=}2$ $\log_3 \left(\dfrac{3}{2}\cdot 6\right)\stackrel{?}{=}2$ $\log_3 9\stackrel{?}{=}2$ $2=2\checkmark$ The solution set is: $\left\{\dfrac{5}{2}\right\}$
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