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: 52



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log x+\log(2x+1)=1 ,$ 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 the Quadratic Formula to solve for the values of the variable. 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(2x+1)]=1 .\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(2x+1)]=1 \\\\ x(2x+1)=10^1 \\\\ 2x^2+x=10 \\\\ 2x^2+x-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 equation above is \begin{array}{l}\require{cancel} (2x+5)(x-2)=0 .\end{array} Equating each factor to zero (Zero Product Property), then the solutions are \begin{array}{l}\require{cancel} 2x+5=0 \\\\\text{OR}\\\\ x-2=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 2x+5=0 \\\\ 2x=-5 \\\\ x=-\dfrac{5}{2} \\\\\text{OR}\\\\ x-2=0 \\\\ x=2 .\end{array} If $ x=-\dfrac{5}{2} ,$ the part of the given expression, $ \log x ,$ becomes $ \log \left(-\dfrac{5}{2} \right) .$ This is not allowed since $ \log x$ is defined only for positive values of $x.$ Hence, only $ x=2 $ satisfies the original equation.
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