College Algebra (11th Edition)

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

Chapter 4 - Review Exercises - Page 469: 59


$x=\left\{ -\dfrac{4}{3}, 5 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log_4[(3x+1)(x-4)]=2 ,$ convert to exponential form. Then use concepts on solving quadratic equations. Do checking of solutions with the original equation. $\bf{\text{Solution Details:}}$ Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to \begin{array}{l}\require{cancel} (3x+1)(x-4)=4^2 \\\\ (3x+1)(x-4)=16 .\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(-4)+1(x)+1(-4)=16 \\\\ 3x^2-12x+x-4=16 \\\\ 3x^2+(-12x+x)+(-4-16)=0 \\\\ 3x^2-11x-20=0 .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the factored form of the equation above is to \begin{array}{l}\require{cancel} (3x+4)(x-5)=0 .\end{array} Equating each factor to zero (Zero Product Property), the solutions are \begin{array}{l}\require{cancel} 3x+4=0 \\\\\text{OR}\\\\ x-5=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 3x+4=0 \\\\ 3x=-4 \\\\ x=-\dfrac{4}{3} \\\\\text{OR}\\\\ x-5=0 \\\\ x=5 .\end{array} Upon checking, both $ x=\left\{ -\dfrac{4}{3}, 5 \right\} $ satisfy the original equation.
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