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: 25


$x=-1 + \sqrt {1+e^4} \approx 6.456$

Work Step by Step

Apply the logarithmic property: $\log_a M+\log_a N = \log_a (MN)$ and rearrange the given expression to obtain: $\ln[(x)(x+2)] \ ...(1)$ Since, $\log_m{n} = 1 $ gives: $m^{(1)}=n$; then we have: $e^4=(x)(x+2)$ $x^2+2x=e^4$ or, $x^2+2x-e^4=0$ This is a quadratic equation; thus by using the quadratic formula $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$, it will become: $x=\dfrac{-(2) \pm \sqrt{(2)^2-(4)(1)(-e^4)}}{2(1)}=$ or, $x=-1 + \sqrt {1+e^4}, -1 - \sqrt {1+e^4}$ Since the domain of the variable is $x \gt 0$, we cannot accept the value of $x=-1 - \sqrt {1+e^4}$ Thus, our answer is: $x=-1 + \sqrt {1+e^4} \approx 6.456$
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