College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.5 - Exponential and Logarithmic Equations - 4.5 Exercises - Page 447: 59



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log_8(x+2)+\log_8(x+4)=\log_88 ,$ 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} \log_8[(x+2)(x+4)]=\log_88 .\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} (x+2)(x+4)=8 .\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} x(x)+x(4)+2(x)+2(4)=8 \\\\ x^2+4x+2x+8=8 \\\\ x^2+6x+8=8 .\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+6x+8-8=0 \\\\ x^2+6x=0 .\end{array} Factoring the $GCF=x,$ the factored form of the equation above is \begin{array}{l}\require{cancel} x(x+6)=0 .\end{array} Equating each factor to zero (Zero Product Property), then the solutions are \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ x+6=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ x+6=0 \\\\ x=-6 .\end{array} If $ x=-6 ,$ the part of the given expression, $ \log_8(x+2) ,$ becomes $ \log_8(-4) .$ This is not allowed since $ \log x$ is defined only for positive values of $x.$ Hence, only $ x=0 $ satisfies the original equation.
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