document.write( "Question 15879: 2log(x+3) - log(x+1) = log(2x+3) What does x equal?? \n" ); document.write( "
Algebra.Com's Answer #7837 by venugopalramana(3286)\"\" \"About 
You can put this solution on YOUR website!
2log(x+3) - log(x+1) = log(2x+3)
\n" ); document.write( "2log(x+3) - log(x+1) - log(2x+3)=0
\n" ); document.write( "using the formula log xy = log x + log y
\n" ); document.write( " and log x/y = log x - log y and log x^n = n*log x , we get
\n" ); document.write( " log (x+3)^2/[(x+1)*(2x+3)] =0 = log (1)
\n" ); document.write( " hence (x+3)^2/[(x+1)*(2x+3)] =1
\n" ); document.write( " (x+3)^2 =(x+1)*(2x+3)
\n" ); document.write( " x^2+6x+9 = 2x^2+5x+3
\n" ); document.write( " 2x^2+5x+3 - (x^2+6x+9) = 0
\n" ); document.write( " 2x^2+5x+3 -x^2-6x-9 = 0
\n" ); document.write( " x^2 -x -6 = 0.......find 2 factors of -6 whose sum is -1....namely -3 and 2
\n" ); document.write( " x^2 +2x - 3x -6 = 0
\n" ); document.write( " x(x+2) - 3 (x+2) = 0
\n" ); document.write( " (x-3)(x+2) = 0.........x-3 = 0 ....Or x+2 =0
\n" ); document.write( " hence x=3 or -2
\n" ); document.write( " \n" ); document.write( "
\n" );