#### Answer

$x=\left\{ 1,\dfrac{10}{3} \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
\log x+\log(13-3x)=1
,$ use the laws of logarithms to simplify the expression. Then change to exponential form. Use 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(13-3x)]=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}
\log_{10}[x(13-3x)]=1
\\\\
x(13-3x)=10^1
\\\\
x(13-3x)=10
.\end{array}
Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
x(13)+x(-3x)=10
\\\\
13x-3x^2=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
\\\\
-1(-3x^2+13x-10)=-1(0)
\\\\
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 expression above is
\begin{array}{l}\require{cancel}
(3x-10)(x-1)=0
.\end{array}
Equating each factor zero (Zero Product Property), the solutions are
\begin{array}{l}\require{cancel}
3x-10=0
\\\\\text{OR}\\\\
x-1=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
3x-10=0
\\\\
3x=10
\\\\
x=\dfrac{10}{3}
\\\\\text{OR}\\\\
x-1=0
\\\\
x=1
.\end{array}
Upon checking, both $
x=\left\{ 1,\dfrac{10}{3} \right\}
$ satisfy the original equation.