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: 63

Answer

$x=25$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log x+\log (x-21)=\log100 ,$ use the properties of logarithms to simplify the left-hand expression. Then drop the logarithm on both sides. 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 equation above is equivalent \begin{array}{l}\require{cancel} \log [x(x-21)]=\log100 .\end{array} Since the logarithmic expressions on both sides have the same base, then the logarithm may be dropped. Hence, the equation above is equivalent to \begin{array}{l}\require{cancel} x(x-21)=100 .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x(x)+x(-21)=100 \\\\ x^2-21x=100 .\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} x^2-21x-100=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-25)(x+4)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x-25=0 \\\\\text{OR}\\\\ x+4=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-25=0 \\\\ x=25 \\\\\text{OR}\\\\ x+4=0 \\\\ x=-4 .\end{array} If $ x=-4 ,$ the part of the given expression, $ \log x ,$ becomes $ \log (-4) .$ This is not allowed since $ \log x$ is defined only for nonnegative values of $x.$ Hence, only $ x=25 $ satisfies the original equation.
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