Precalculus (6th Edition) Blitzer

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

Chapter 3 - Section 3.4 - Exponential and Logarithmic Equations - Exercise Set - Page 489: 87

Answer

Solution set = $\{2\}.$

Work Step by Step

For the equation to be defined, logarithms must have positive arguments $\left[\begin{array}{llll} x \gt 0, & and & x+3 \gt 0 & \\ & & x\gt-3 & \end{array}\right]$ $ x \gt -3 \qquad (*)$ This is the condition eventual solutions must satisfy. $\log x+\log(x+3)=\log 10\qquad $... LHS: product rule $\log(x^{2}+3x)=\log 10 \quad $ ... $\log $ is one-to-one $ x^{2}+3x=10$ $ x^{2}+3x-10=0$ Factor: +5 and -2 are factors of -10 whose sum is +3. $(x+5)(x-2)=0$ $ x=2 \qquad $ ... satisfies (*) $ x=-5 \quad $ ... does not satisfy (*) Thus, the solution set = $\{2\}.$
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