#### Answer

$x=5$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
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.