College Algebra (11th Edition)

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

Chapter 4 - Test - Page 472: 19



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log_2 x+ \log_2(x+2)=3 ,$ use the properties of logarithms to simplify the expression at the left. Then convert to exponential form and use the concepts of solving quadratic equations. Do checking of solution/s 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_2 [x(x+2)]=3 .\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} x(x+2)=2^3 \\\\ x(x+2)=8 .\end{array} In the form $ax^2+bx+c=0,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x(x)+x(2)=8 \\\\ x^2+2x=8 \\\\ x^2+2x-8=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+4)(x-2)=0 .\end{array} Equating each factor to zero (Zero Product Property), the solutions are \begin{array}{l}\require{cancel} x+4=0 \\\\\text{OR}\\\\ x-2=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x+4=0 \\\\ x=-4 \\\\\text{OR}\\\\ x-2=0 \\\\ x=2 .\end{array} If $x=-4,$ the part of the given expression ,$ \log_2 x ,$ becomes $ \log_2 -4 ,$ which is not a real number. Hence, oly $ x=2 ,$ satisfies the original equation.
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