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: 45

Answer

$x \approx 4.405$

Work Step by Step

$\log_3 (x+2) + \log_3 (x-3) = 2$ $\log_3 (x+2)(x-3) = 2$ $3^{2} = (x+2)(x-3)$ $9 = (x+2)(x-3)$ $9 = x(x-3)+2(x-3)$ $9 = x^{2} - 3x + 2x - 6$ $x^{2} -x - 6 - 9 = 0$ $x^{2} - x - 15 = 0$ $x = \frac{-(-1)±\sqrt {(-1)^{2}-4(1)(-15)}}{2(1)}$ $x = \frac{1±\sqrt {1-4(1)(-15)}}{2}$ $x = \frac{1±\sqrt {1+60}}{2}$ $x = \frac{1±\sqrt {61}}{2}$ $x \approx 4.405, -3.405$ Since $x$ can't be negative, the only possible solution is $x = 4.405$. This is because $\log$ of a negative number does not exist and therefore we only accept the positive number of $x$. Check: $\log_3 ((\frac{1+\sqrt {61}}{2})+2) + \log_3 ((\frac{1+\sqrt {61}}{2})-3) \overset{?}{=} 2$ $\log_3 (6.405124...) + \log_3 (1.405124...) \overset{?}{=} 2$ $\log_3 [(6.405124...) \times (1.405124...)] \overset{?}{=} 2$ $\log_3 9\overset{?}{=} 2$ $\log_3 3^{2}\overset{?}{=} 2$ $2\log_3 3\overset{?}{=} 2$ $2(1)\overset{?}{=} 2$ $2 = 2$
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