Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 6 - Logarithmic Functions - 6.6 Solving Logarithmic Equations - 6.6 Exercises: 65

Answer

$x = 1, -2.500$

Work Step by Step

$4x^{2} + 6x - 10 = 0$ $2(2x^{2} + 3x - 5) = 0$ $2(2x^{2} + 5x - 2x - 5) = 0$ $2(x(2x+5) - 1(2x + 5)) = 0$ $2(x-1)(2x+5) = 0$ $x = 1, -\frac{5}{2}$ $x = 1, -2.500$ $2x + 5 = 0$ $2x = -5$ $x = -\frac{5}{2}$ Check: When $x = 1$ $4x^{2} + 6x - 10 \overset{?}{=} 0$ $4(1)^{2} + 6(1) - 10 \overset{?}{=} 0$ $4 + 6 - 10 \overset{?}{=} 0$ $10 - 10 \overset{?}{=} 0$ $0 = 0$ When $x = -\frac{5}{2}$ $4x^{2} + 6x - 10 \overset{?}{=} 0$ $4(-\frac{5}{2})^{2} + 6(-\frac{5}{2}) - 10 \overset{?}{=} 0$ $4(6.25) - 15 - 10 \overset{?}{=} 0$ $25 - 15 - 10 \overset{?}{=} 0$ $25 - 25 \overset{?}{=} 0$ $0 = 0$
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