## College Algebra (11th Edition)

$x=\dfrac{5}{2}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\ln(4x-2)-\ln4=-\ln(x-2) ,$ use the properties of logarithms to combine the logarithmic expressions. 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 properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} \ln(4x-2)+\ln(x-2)=\ln4 .\end{array} Using the Product Rule of Logarithms, which is given by $\log_b (xy)=\log_bx+\log_by,$ the equation above is equivalent \begin{array}{l}\require{cancel} \ln[(4x-2)(x-2)]=\ln4 .\end{array} Since the logarithmic expressions on both sides have the same base, then the logarithm may be dropped. Hence, the equation above is equivalent to \begin{array}{l}\require{cancel} (4x-2)(x-2)=4 .\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} 4x(x)+4x(-2)-2(x)-2(-2)=4 \\\\ 4x^2-8x-2x+4=4 \\\\ 4x^2+(-8x-2x)+(4-4)=0 \\\\ 4x^2-10x=0 .\end{array} Factoring the $GCF=2x,$ the factored form of the equation above is \begin{array}{l}\require{cancel} 2x(2x-5)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} 2x=0 \\\\\text{OR}\\\\ 2x-5=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 2x=0 \\\\ x=\dfrac{0}{2} \\\\ x=0 \\\\\text{OR}\\\\ 2x-5=0 \\\\ 2x=5 \\\\ x=\dfrac{5}{2} .\end{array} If $x=0 ,$ the part of the given expression, $\ln(4x-2) ,$ becomes $\ln(-2) .$ This is not allowed since $\ln x$ is defined only for nonnegative values of $x.$ Hence, only $x=\dfrac{5}{2}$ satisfies the original equation.