Answer
$x=3$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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.