document.write( "Question 20619: Here is my question:\r
\n" ); document.write( "\n" ); document.write( "1-logx = log(3x-1)\r
\n" ); document.write( "\n" ); document.write( "I started with:
\n" ); document.write( "1 - logx = log3x - log
\n" ); document.write( "1 = log4x - log\r
\n" ); document.write( "\n" ); document.write( "The answer is 2, but I don't know where to go from here!
\n" ); document.write( "Can anyone help?
\n" ); document.write( "Thanks!
\n" ); document.write( "Sandy
\n" ); document.write( "

Algebra.Com's Answer #14078 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
1-logx=log(3x-1)
\n" ); document.write( "1=logx+log(3x-1)
\n" ); document.write( "You have to know log(AB)=logA+logB
\n" ); document.write( "So, 1=log[x(3x-1)]
\n" ); document.write( "1=log[3x^2-x]
\n" ); document.write( "Assuming the base is 10 you get:
\n" ); document.write( "10^1=3x^2-x
\n" ); document.write( "3x^2-x-10=0
\n" ); document.write( "(3x+5)(x-2)=0
\n" ); document.write( "x cannot be negative so x=2\r
\n" ); document.write( "\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
\n" ); document.write( "
\n" );