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: 70

Answer

$x=8 $

Work Step by Step

$\bf{\text{Solution Outline:}}$ First, use the properties of logarithms to simplify the given logarithmic expression, $ \log_2(x-7)+\log_2 x=3 ,$ to a single logarithm. Then convert to exponential form. Next, use concepts of quadratic equations to solve for the variable. Finally, it is a must to do checking at the end. $\bf{\text{Solution Details:}}$ Using $\log_b (xy)=\log_b x+\log_b y$ or the Product Rule of logarithms, the given expression is equivalent to \begin{array}{l}\require{cancel} \log_2[(x-7)x]=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-7)x=2^3 \\\\ (x-7)x=8 .\end{array} Using the properties of equality and the concepts of quadratic equations, the solution/s to the equation above is/are \begin{array}{l}\require{cancel} x^2-7x=8 \\\\ x^2-7x-8=0 \\\\ (x-8)(x+1)=0 .\end{array} Equating each factor to $0$ and then solving for the variable, then the solutions are $ x=\left\{ -1,8 \right\} .$ Upon checking, only $ x=8 $ 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.