This preview shows page 1. Sign up to view the full content.
Unformatted text preview: s of zero so this is a solution. x = 2 : log ( 2 ) + log ( 2  1) = log ( 3 ( 2 ) + 12 ) We don’t need to go any farther, there is a logarithm of a negative number in the first term (the
others are also negative) and that’s all we need in order to exclude this as a solution.
Be careful here. We are not excluding x = 2 because it is negative, that’s not the problem. We
are excluding it because once we plug it into the original equation we end up with logarithms of
negative numbers. It is possible to have negative values of x be solutions to these problems, so
don’t mistake the reason for excluding this value. © 2007 Paul Dawkins 303 http://tutorial.math.lamar.edu/terms.aspx College Algebra Also, along those lines we didn’t take x = 6 as a solution because it was positive, but because it
didn’t produce any negative numbers or zero in the logarithms upon substitution. It is possible
for negative numbers to not be solutions.
So, with all that out of the way, we’ve got a single solution to this equation, x = 6 .
[Return to Problems...
View
Full
Document
This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.
 Spring '12
 MrVinh

Click to edit the document details