Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 11 - Cumulative Review Exercises - Page 872: 8


The solution set is$\left\{ 9 \right\}$.

Work Step by Step

A logarithmic equation in one variable can be solved when the equation is written in exponential form. For example, the equation \[{{\log }_{a}}x=b\] when written in exponential form is\[x={{a}^{b}}\]. Hence, solution for x is \[x={{a}^{b}}\]. Now, for the provided logarithmic equation, simplify by logarithmic properties: \[\begin{align} & {{\log }_{9}}\left( x \right)+{{\log }_{9}}\left( x-8 \right)=1 \\ & {{\log }_{9}}\left( x\left( x-8 \right) \right)=1 \end{align}\] Now, rewrite the logarithmic equation in the exponential form: \[\begin{align} & x\left( x-8 \right)=9 \\ & {{x}^{2}}-8x-9=0 \\ & {{x}^{2}}-9x+x-9=0 \\ & x\left( x-9 \right)+1\left( x-9 \right)=0 \end{align}\] Solve further as: \[\left( x+1 \right)\left( x-9 \right)=0\] Case 1: \[x=-1\] Argument of \[{{\log }_{9}}\left( x \right)\] is negative in this case which is not possible. Case 2: \[x=9\] Arguments of both logs are positive in this case. Hence, the solution of the equation is \[x=9\].
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.