## College Algebra (11th Edition)

$x=3$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\log_2(x-2)+\log_2(x-1)=1 ,$ use the properties of logarithms to simplify the left-hand expression. Then convert to exponential form. 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} \log_2[(x-2)(x-1)]=1 .\end{array} Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to \begin{array}{l}\require{cancel} (x-2)(x-1)=2^1 \\\\ (x-2)(x-1)=2 .\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} x(x)+x(-1)-2(x)-2(-1)=2 \\\\ x^2-x-2x+2=2 \\\\ x^2-3x+2=2 .\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^2-3x+2-2=0 \\\\ x^2-3x=0 .\end{array} Factoring the $GCF=x,$ the factored form of the equation above is \begin{array}{l}\require{cancel} x(x-3)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ x-3=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ x-3=0 \\\\ x=3 .\end{array} If $x=0 ,$ the part of the given expression, $\log_2(x-2) ,$ becomes $\log_2(-2) .$ This is not allowed since $\log x$ is defined only for nonnegative values of $x.$ Hence, only $x=3$ satisfies the original equation.