College Algebra (11th Edition)

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

Chapter 4 - Review Exercises - Page 469: 61


$x=\left\{ 1,\dfrac{10}{3} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log x+\log(13-3x)=1 ,$ use the laws of logarithms to simplify the expression. Then change to exponential form. Use 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(13-3x)]=1 .\end{array} Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to \begin{array}{l}\require{cancel} \log_{10}[x(13-3x)]=1 \\\\ x(13-3x)=10^1 \\\\ x(13-3x)=10 .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} x(13)+x(-3x)=10 \\\\ 13x-3x^2=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 \\\\ -1(-3x^2+13x-10)=-1(0) \\\\ 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 expression above is \begin{array}{l}\require{cancel} (3x-10)(x-1)=0 .\end{array} Equating each factor zero (Zero Product Property), the solutions are \begin{array}{l}\require{cancel} 3x-10=0 \\\\\text{OR}\\\\ x-1=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 3x-10=0 \\\\ 3x=10 \\\\ x=\dfrac{10}{3} \\\\\text{OR}\\\\ x-1=0 \\\\ x=1 .\end{array} Upon checking, both $ x=\left\{ 1,\dfrac{10}{3} \right\} $ satisfy the original equation.
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