College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.5 - Exponential and Logarithmic Equations - 4.5 Exercises - Page 447: 48


$x=\left\{ -\dfrac{14}{3},8 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log_4[(3x+8)(x-6)]=3 ,$ change to exponential form. Then express the resulting equation in the form $ax^2+bx+c=0.$ Use the Quadratic Formula to solve for the values of the variable. Then do checking of the solutions with the original equation. $\bf{\text{Solution Details:}}$ Since $y=b^x$ is equivalent to $\log_b y=x,$ the exponential form of the equation above is \begin{array}{l}\require{cancel} (3x+8)(x-6)=4^3 \\\\ (3x+8)(x-6)=64 .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} 3x(x)+3x(-6)+8(x)+8(-6)=64 \\\\ 3x^2-18x+8x-48=64 \\\\ 3x^2+(-18x+8x)+(-48-64)=0 \\\\ 3x^2-10x-112=0 .\end{array} In the equation above, $a= 3 ,$ $b= -10 ,$ and $c= -112 .$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then \begin{array}{l}\require{cancel} x=\dfrac{-(-10)\pm\sqrt{(-10)^2-4(3)(-112)}}{2(3)} \\\\ x=\dfrac{10\pm\sqrt{100+1344}}{6} \\\\ x=\dfrac{10\pm\sqrt{1444}}{6} \\\\ x=\dfrac{10\pm\sqrt{(38)^2}}{6} \\\\ x=\dfrac{10\pm38}{6} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{10-38}{6} \\\\ x=\dfrac{-28}{6} \\\\ x=\dfrac{-14}{3} \\\\ x=-\dfrac{14}{3} \\\\\text{OR}\\\\ x=\dfrac{10+38}{6} \\\\ x=\dfrac{48}{6} \\\\ x=8 .\end{array} Upon checking, $ x=\left\{ -\dfrac{14}{3},8 \right\} ,$ satisfy the original equation.
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