#### Answer

$x=-13$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
\ln(10-x)+\ln(-6-x)=\ln(-34-15x)
,$ 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 expression above is equivalent \begin{array}{l}\require{cancel} \ln[(10-x)(-6-x)]=\ln(-34-15x) .\end{array}
Since the logarithm on both sides have the same base, then the logarithm can be dropped. Hence, the equation above is equivalent to \begin{array}{l}\require{cancel} (10-x)(-6-x)=-34-15x .\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} 10(-6)+10(-x)-x(-6)-x(-x)=-34-15x \\\\ -60-10x+6x+x^2=-34-15x \\\\ -60-4x+x^2=-34-15x .\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+(-4x+15x)+(-60+34)=0 \\\\ x^2+11x-26=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}
(x+13)(x-2)=0
.\end{array}
Equating each factor to zero (Zero Product Property), then the solutions are
\begin{array}{l}\require{cancel}
x+13=0
\\\\\text{OR}\\\\
x-2=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x+13=0
\\\\
x=-13
\\\\\text{OR}\\\\
x-2=0
\\\\
x=2
.\end{array}
If $
x=2
,$ the part of the given expression, $
\ln(-6-x)
,$ becomes $
\ln(-8)
.$ This is not allowed since $
\ln x$ is defined only for positive values of $x.$ Hence, only $
x=-13
$ satisfies the original equation.