document.write( "Question 199628: what values of x cannot possibly be solutions of the following equation?
\n" ); document.write( "LOGa(3x+1)=2
\n" ); document.write( "

Algebra.Com's Answer #150020 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
Recall that you CANNOT evaluate the log of a negative value or 0. So this means that \"3x%2B1%3E0\"\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"3x%2B1%3E0\" Start with the given inequality.\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"3x%3E0-1\" Subtract \"1\" from both sides.\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"3x%3E-1\" Combine like terms on the right side.\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"x%3E-1%2F3\" Divide both sides by \"3\" to isolate \"x\". \r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "This means that any values of x that are either equal to \"-1%2F3\" or are less than \"-1%2F3\" are NOT allowed.
\n" ); document.write( "
\n" );