## College Algebra (11th Edition)

$x=\left\{ -\dfrac{4}{3}, 5 \right\}$
$\bf{\text{Solution Outline:}}$ 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.