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: 72



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log_5(3x+2)+\log_5(x-1)=1 ,$ use the properties of logarithms to simplify the logarithmic expressions on the left side. Then change to exponential form. 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_5[(3x+2)(x-1)]=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} (3x+2)(x-1)=5^1 \\\\ (3x+2)(x-1)=5 .\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} 3x(x)+3x(-1)+2(x)+2(-1)=5 \\\\ 3x^2-3x+2x-2=5 \\\\ 3x^2+(-3x+2x)+(-2-5)=0 \\\\ 3x^2-x-7=0 .\end{array} In the equation above, $a= 3 ,$ $b= -1 ,$ and $c= -7 .$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then \begin{array}{l}\require{cancel} x=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(3)(-7)}}{2(3)} \\\\ x=\dfrac{1\pm\sqrt{1+84}}{6} \\\\ x=\dfrac{1\pm\sqrt{85}}{6} .\end{array} Upon checking, only $ x=\dfrac{1+\sqrt{85}}{6} $ satisfies the original equation.
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