## Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole

# Chapter 1-6 - Cumulative Review - Page 547: 11

#### Answer

$x = \frac{5+\sqrt {481}}{2} \approx 13.466$

#### Work Step by Step

$\log (x+2) + \log(x-7) = 2$ $\log(x+2)(x-7) = 2$ $\log_{10} (x+2)(x-7) = 2$ $10^{2} = (x+2)(x-7)$ $100 = (x+2)(x-7)$ $x(x-7) +2 (x-7) = 100$ $x^{2} - 7x + 2x - 14 = 100$ $x^{2} - 5x - 14 - 100 = 0$ $x^{2} - 5x - 114 = 0$ $x = \frac{-(-5)±\sqrt {(-5)^{2}-4(1)(-114)}}{2(1)}$ $x = \frac{5±\sqrt {25-4(1)(-114)}}{2}$ $x = \frac{5±\sqrt {25+456}}{2}$ $x = \frac{5±\sqrt {481}}{2}$ Since we can't take the log of a negative number, the only possible solution is $x = \frac{5+\sqrt {481}}{2}$. $x = \frac{5+\sqrt {481}}{2} \approx 13.466$ Check: $\log ((13.466)+2) + \log(6.465...) \overset{?}{=}2$ $\log (15.4658...) + \log(6.4658...) \overset{?}{=} 2$ $\log (15.4658...)(6.4658...) \overset{?}{=} 2$ $\log 100 \overset{?}{=} 2$ $\log_{10} 10^{2} \overset{?}{=} 2$ $2(\log_{10} 10) \overset{?}{=} 2$ $2(1) \overset{?}{=} 2$ $2 = 2$

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