## College Algebra (11th Edition)

Published by Pearson

# Chapter 4 - Section 4.5 - Exponential and Logarithmic Equations - 4.5 Exercises - Page 447: 51

#### Answer

$x=5$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $\log x+\log(x+15)=2 ,$ use the properties of logarithms to simplify the left-hand expression. Then change to exponential form. Express the resulting equation in the form $ax^2+bx+c=0$ and use concepts of solving quadratic equations. Then 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 x(x+15)=2 .\end{array} Since $y=b^x$ is equivalent to $\log_b y=x,$ the exponential form of the equation above is \begin{array}{l}\require{cancel} \log_{10} x(x+15)=2 \\\\ x(x+15)=10^2 \\\\ x^2+15x=100 \\\\ x^2+15x-100=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+20)(x-5)=0 .\end{array} Equating each factor to zero (Zero Product Property), then the solutions are \begin{array}{l}\require{cancel} x+20=0 \\\\\text{OR}\\\\ x=5 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x+20=0 \\\\ x=-20 \\\\\text{OR}\\\\ x-5=0 \\\\ x=5 .\end{array} If $x=-20 ,$ the part of the given expression, $\log x ,$ becomes $\log (-20) .$ This is not allowed since $\log x$ is defined only for positive values of $x.$ Hence, only $x=5$ satisfies the original equation.

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