Precalculus (6th Edition) Blitzer

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

Chapter 3 - Review Exercises - Page 513: 77


The value of the given expression has no solution.

Work Step by Step

Consider the given equation, ${{\log }_{3}}\left( x-1 \right)-{{\log }_{3}}\left( x+2 \right)=2$ Apply the quotient rule of logarithms: $\begin{align} & {{\log }_{3}}\left( x-1 \right)-{{\log }_{3}}\left( x+2 \right)=2 \\ & {{\log }_{3}}\frac{\left( x-1 \right)}{\left( x+2 \right)}=2 \end{align}$ The provided equation can be written as $\begin{align} & \frac{\left( x-1 \right)}{\left( x+2 \right)}={{3}^{2}} \\ & \frac{\left( x-1 \right)}{\left( x+2 \right)}=9 \\ & \left( x-1 \right)=9\left( x+2 \right) \\ \end{align}$ $ x-1=9x+18$ Now, add to both sides $-x-18$: $\begin{align} & x-1-x-18=9x+18-x-18 \\ & 8x=-19 \\ & x=\frac{-19}{8} \\ \end{align}$ Here $ x=\frac{-19}{8}$ is not the solution of the given expression because $\begin{align} & {{\log }_{3}}\left( x-1 \right)-{{\log }_{3}}\left( x+2 \right)=2 \\ & {{\log }_{3}}\left( \frac{-19}{8}-1 \right)-{{\log }_{3}}\left( \frac{-19}{8}+2 \right)=2 \\ & {{\log }_{3}}\left( \frac{-19}{8}-1 \right)-{{\log }_{3}}\left( \frac{-19}{8}+2 \right)=2 \\ & {{\log }_{3}}\left( -3.375 \right)-{{\log }_{3}}\left( -0.375 \right)=2 \end{align}$ By definition of ${{\log }_{a}}b $, $ b>0$ but $-3.375$ and $-0.375$ is not greater than zero. Thus, $ x=\frac{-19}{8}$ does not satisfy the equation ${{\log }_{3}}\left( x-1 \right)-{{\log }_{3}}\left( x+2 \right)=2$ Therefore, there is no solution of the given expression.
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