Therefore we get a single solution for this equation

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 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.

Ask a homework question - tutors are online