College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Section 6.6 - Logarithmic and Exponential Equations - 6.6 Assess Your Understanding - Page 466: 84

Answer

$x=\dfrac{26-8\sqrt{10}}{9}\approx0.078$ and $x=\dfrac{26+8\sqrt{10}}{9}\approx5.700$

Work Step by Step

Recall: (1) The Change-of-Base Formula for Logarithms: $\log_a{b}=\dfrac{\log_c{b}}{\log_c{a}}$ (2) $\log_a{a^b}=b$ (3) $n\cdot \log_a{b} = \log_a{(b^n)}$ (4) $\log_a{b} - \log_a{c} = \log_a{\left(\frac{b}{c}\right)}$ Use the change-of-base formula above to obtain: \begin{align*} \log_2{(3x+2)}-\dfrac{\log_2{x}}{\log_2{4}}&=3\\ \log_2{(3x+2)}-\dfrac{\log_2{x}}{\log_2{(2^2)}}&=3\\ \end{align*} Use the rule in (2) above to obtain: \begin{align*} \log_2{(3x+2)}-\dfrac{\log_2{x}}{2}&=3\\ 2\left(\log_2{(3x+2)}-\dfrac{\log_2{x}}{2}\right)&=3(2)\\ 2\log_2{(3x+2)}-\log_2{x}&=6\\ \end{align*} Use the rules in (3) and (4) above to obtain: \begin{align*} \log_2{(3x+2)^2}-\log_2{x}&=6\\ \log_2{\left(\frac{(3x+2)^2}{x}\right)}&=6 \end{align*} Use the rule $\log_a{b}=y \longrightarrow a^y=b$ to obtain: \begin{align*} \frac{(3x+2)^2}{x}&=2^6\\ \frac{(3x)^2+2(3x)(2)+2^2}{x}&=64\\ \frac{9x^2+12x+4}{x}&=64\\ x\left(\frac{9x^2+12x+4}{x}\right)&=64(x)\\ 9x^2+12x+4&=64x\\ 9x^2+12x+4-64x&=0\\ 9x^2-52x+4&=0 \end{align*} Solve the quadratic equation using the quadratic formula $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ with $a=9, b=-52, \text{ and } c=4$ to obtain: \begin{align*} x&=\frac{-(-52)\pm\sqrt{(-52)^2-4(9)(4)}}{2(9)}\\ &=\frac{52\pm\sqrt{2704-144}}{18}\\ &=\frac{52\pm\sqrt{2560}}{18}\\ &=\frac{52\pm\sqrt{256(10)}}{18}\\ &=\frac{52\pm\sqrt{16^2(10)}}{18}\\ &=\frac{52\pm16\sqrt{10}}{18}\\ &=\frac{26\pm8\sqrt{10}}{9} \end{align*} Therefore, the solutions are: $x=\dfrac{26-8\sqrt{10}}{9}\approx0.078$ and $x=\dfrac{26+8\sqrt{10}}{9}\approx5.700$
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