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: 27


$x=\dfrac{-5+3 \sqrt{5}}{2} \approx 0.854$

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_3[(x+1)(x+4)] $ Since, $\log_m{n} = 1 $ gives: $m^{(1)}=n$, then we have: $\log_3(x^2+5x+4)=2$ $x^2+5x+4=3^2$ or, $x^2+5x-5=0$ This is a quadratic equation; thus by using the quadratic formula $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$, we get: $x=\dfrac{-5 \pm 3 \sqrt{5}}{2(1)}$ or, $x=\dfrac{-5 + 3 \sqrt{5}}{2}, \dfrac{-5 -3 \sqrt{5}}{2}$ Since, the domain of the variable is $x \gt 0$, we cannot accept the value of $x=\dfrac{-5 -3 \sqrt{5}}{2}$ Thus, our answer is: $x=\dfrac{-5+3 \sqrt{5}}{2} \approx 0.854$
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