# Chapter 4 - Section 4.5 - Exponential and Logarithmic Equations - 4.5 Exercises - Page 447: 58

$x=-13$

#### 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.

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