Precalculus (6th Edition) Blitzer

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

Chapter 8 - Cumulative Review Exercises - Page 952: 6


The solution is $x=\,\,1$.

Work Step by Step

Consider the given equation system, $\begin{align} & {{\log }_{3}}x+{{\log }_{3}}\left( x+2 \right)=1 \\ & {{\log }_{3}}\left\{ x\left( x+2 \right) \right\}=1 \\ & x\left( x+2 \right)={{3}^{1}} \end{align}$ It can be further simplified as below: $\begin{align} & {{x}^{2}}+2x-3=0 \\ & {{x}^{2}}+3x-x-3=0 \\ & x\left( x+3 \right)-\left( x+3 \right)=0 \\ & \left( x+3 \right)\left( x-1 \right)=0 \end{align}$ Therefore $x=-3\,\,\text{or}\,\,1$ ${{\log }_{a}}x=b$ can also be written as ${{a}^{b}}=x$, where $a>0$ so for any value of b, x is always positive. Therefore, discard the negative value of the logarithm. The solutions of the equations are $x=\,\,1$.
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