Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Section 4.6 Logarithmic and Exponential Equations - 4.6 Assess Your Understanding - Page 337: 28

Answer

$x=- 4 +\sqrt{17}\approx 0.123$

Work Step by Step

Apply the logarithmic property: $\log_a M+\log_a N = \log_a (MN)$ and rearrange the given expression to obtain: $\log_2[(x+1)(x+7)] $ Since $\log_m{n} = 1 $ gives: $m^{(1)}=n$, then we have: $\log_2(x^2+8x+7)=3$ $x^2+8x+7=2^3$ or, $x^2+8x-1=0$ This is a quadratic equation; thus by using the quadratic formula $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$, it will become: $x=\dfrac{-8 \pm \sqrt{8^2-(4)(1)(1)}}{2(1)}$ or, $x=-4 + \sqrt{17}, - 4 - \sqrt{17}$ Since, the domain of the variable is $x \gt 0$, we cannot accept the value of $x=- 4 - \sqrt{17}$ Thus, our answer is: $x=- 4 +\sqrt{17}\approx 0.123$
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