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ln(x+2)=ln(2x-1)+3
\n" ); document.write( "ln(x+2)-ln(2x-1) = 3----(1) (here the base is 10 )
\n" ); document.write( "log[(x+2)/(2x-1)] =3 (using log(a)-log(b) =log(a/b) )
\n" ); document.write( "(x+2)/(2x-1)=10^3
\n" ); document.write( "(x+2) = (2x-1)(10^3)
\n" ); document.write( "x+2 = 2X(10)^3(x)-(10^3)
\n" ); document.write( "x+2 = 2000x-1000
\n" ); document.write( "2+1000=2000x-x
\n" ); document.write( "1002=1999x
\n" ); document.write( "Therefore x = 1002/1999
\n" ); document.write( "Answer: x = 1002/1999
\n" ); document.write( "Verification: Putting x = 1002/1999 in (1)
\n" ); document.write( "LHS = ln(x+2)-ln(2x-1)
\n" ); document.write( "=log[(1002/1999)+2] - log[(2004/1999)-1]
\n" ); document.write( "=log{[1002+(2X1999)]/1999}- log{[(2004-1999)]/1999}
\n" ); document.write( "=log{[1002+3998]/1999}- log{(5)/1999}
\n" ); document.write( "=log{(5000)/1999}- log{(5)/1999}
\n" ); document.write( "=(log5000-log1999)-(log5-log1999)
\n" ); document.write( "=log5000-log1999-log5+log1999
\n" ); document.write( "=log5000-log5
\n" ); document.write( "=log(5000/5)
\n" ); document.write( "=log(1000)
\n" ); document.write( "=log[(10)^3]
\n" ); document.write( "=3Xlog(10)
\n" ); document.write( "=3X1 (since log10 to the same base 10 is 1)
\n" ); document.write( "=3 = RHS \n" ); document.write( "