Precalculus (6th Edition) Blitzer

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

Chapter 7 - Cumulative Review Exercises - Page 880: 17


The solution is $\left\{ 2 \right\}$.

Work Step by Step

Solve the provided equation as follows: $\begin{align} & \log \left( x+3 \right)+\log x=1 \\ & log\left[ \left( x+3 \right)\left( 3 \right) \right]=1 \\ & \left( x+3 \right)x={{10}^{1}} \\ & {{x}^{2}}+3x=10 \end{align}$ Further solve this equation $\begin{align} & {{x}^{2}}+3x-10=0 \\ & {{x}^{2}}+5x-2x-10=0 \\ & x\left( x+5 \right)-2\left( x+5 \right)=0 \\ & \left( x+5 \right)\left( x-2 \right)=0 \end{align}$ Finally, we get, $ x=-5,2$ Put the value of $ x=-5$ in the original equation, $\log \left( -5+3 \right)+\log \left( -5 \right)=1$ Since, log of a negative number is not defined, $ x=-5$ is not solution Thus the solution set is $\left\{ 2 \right\}$.
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