Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.1 - Systems of Linear Equations in Two Variables - Exercise Set - Page 823: 113


The solution is $\left\{ 3 \right\}$.

Work Step by Step

$\begin{align} & {{\log }_{3}}\left[ x\left( x+6 \right) \right]=3 \\ & x\left( x+6 \right)=\text{anti}{{\log }_{3}}\left( 3 \right) \\ & x\left( x+6 \right)={{3}^{3}} \end{align}$ Solving further, we get, $\begin{align} & {{x}^{2}}+6x=27 \\ & {{x}^{2}}+6x-27=0 \\ & {{x}^{2}}+9x-3x-27=0 \\ & \left( x+9 \right)\left( x-3 \right)=0 \end{align}$ Therefore, $x=-9,3$ Put $x=-9$ in the original equation. ${{\log }_{3}}\left( -9 \right)+{{\log }_{3}}\left( -9+6 \right)=3$ The log of a negative number is not defined. Thus, $x=-9$ is an extraneous solution. Thus, $\left\{ 3 \right\}$ is the only solution.
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