## College Algebra (11th Edition)

$x=2$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\log x+\log(2x+1)=1 ,$ use the properties of logarithms to simplify the left-hand expression. Then change to exponential form. Express the resulting equation in the form $ax^2+bx+c=0$ and use the Quadratic Formula to solve for the values of the variable. Then 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} \log [x(2x+1)]=1 .\end{array} Since $y=b^x$ is equivalent to $\log_b y=x,$ the exponential form of the equation above is \begin{array}{l}\require{cancel} \log_{10} [x(2x+1)]=1 \\\\ x(2x+1)=10^1 \\\\ 2x^2+x=10 \\\\ 2x^2+x-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} (2x+5)(x-2)=0 .\end{array} Equating each factor to zero (Zero Product Property), then the solutions are \begin{array}{l}\require{cancel} 2x+5=0 \\\\\text{OR}\\\\ x-2=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 2x+5=0 \\\\ 2x=-5 \\\\ x=-\dfrac{5}{2} \\\\\text{OR}\\\\ x-2=0 \\\\ x=2 .\end{array} If $x=-\dfrac{5}{2} ,$ the part of the given expression, $\log x ,$ becomes $\log \left(-\dfrac{5}{2} \right) .$ This is not allowed since $\log x$ is defined only for positive values of $x.$ Hence, only $x=2$ satisfies the original equation.