College Algebra (11th Edition)

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

Chapter 4 - Test: 19

Answer

$x=2$

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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.