## College Algebra (11th Edition)

$x=-8$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\ln(5-x)+\ln(-3-x)=\ln(1-8x) ,$ use the properties of logarithms to simplify the left-hand expression. Then drop the logarithm on both sides. Use the properties of equality and concepts on solving quadratic equations. Finally, do checking of the solutions with the original equation. $\bf{\text{Solution Details:}}$ Using the Product Rule of Logarithms, which is given by $\log_b (xy)=\log_bx+\log_by,$ the expression above is equivalent \begin{array}{l}\require{cancel} \ln[(5-x)(-3-x)]=\ln(1-8x) .\end{array} Since the logarithm on both sides have the same base, then the logarithm can be dropped. Hence, the equation above is equivalent to \begin{array}{l}\require{cancel} (5-x)(-3-x)=1-8x .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the equation above is equivalent to \begin{array}{l}\require{cancel} 5(-3)+5(-x)-x(-3)-x(-x)=1-8x \\\\ -15-5x+3x+x^2=1-8x \\\\ -15-2x+x^2=1-8x .\end{array} Using the properties of equality, in the form $ax^2+bx+c=0,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x^2+(-2x+8x)+(-15-1)=0 \\\\ x^2+6x-16=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 \begin{array}{l}\require{cancel} (x+8)(x-2)=0 .\end{array} Equating each factor to zero (Zero Product Property), then the solutions are \begin{array}{l}\require{cancel} x+8=0 \\\\\text{OR}\\\\ x-2=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x+8=0 \\\\ x=-8 \\\\\text{OR}\\\\ x-2=0 \\\\ x=2 .\end{array} If $x=2 ,$ the part of the given expression, $\ln(-3-x) ,$ becomes $\ln(-5) .$ This is not allowed since $\ln x$ is defined only for positive values of $x.$ Hence, only $x=-8$ satisfies the original equation.