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: 51

Answer

$x=5$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log x+\log(x+15)=2 ,$ 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 concepts of solving quadratic equations. 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(x+15)=2 .\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(x+15)=2 \\\\ x(x+15)=10^2 \\\\ x^2+15x=100 \\\\ x^2+15x-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+20)(x-5)=0 .\end{array} Equating each factor to zero (Zero Product Property), then the solutions are \begin{array}{l}\require{cancel} x+20=0 \\\\\text{OR}\\\\ x=5 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x+20=0 \\\\ x=-20 \\\\\text{OR}\\\\ x-5=0 \\\\ x=5 .\end{array} If $ x=-20 ,$ the part of the given expression, $ \log x ,$ becomes $ \log (-20) .$ This is not allowed since $ \log x$ is defined only for positive values of $x.$ Hence, only $ x=5 $ 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.