College Algebra (11th Edition)

$x=5$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\log x+\log (3x-13)=\log10 ,$ 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 equation above is equivalent \begin{array}{l}\require{cancel} \log [x(3x-13)]=\log10 .\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} x(3x-13)=10 .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x(3x)+x(-13)=10 \\\\ 3x^2-13x=10 .\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} 3x^2-13x-10=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} (3x+2)(x-5)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} 3x+2=0 \\\\\text{OR}\\\\ x-5=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 3x+2=0 \\\\ 3x=-2 \\\\ x=-\dfrac{2}{3} \\\\\text{OR}\\\\ x-5=0 \\\\ x=5 .\end{array} If $x=-\dfrac{2}{3} ,$ the part of the given expression, $\log x ,$ becomes $\log \left(-\dfrac{2}{3}\right) .$ This is not allowed since $\log x$ is defined only for nonnegative values of $x.$ Hence, only $x=5$ satisfies the original equation.