document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1239>Question 1239</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve ln(3x+8)=ln(2x+2)+ln(x-2) for x\r<BR>\n" );
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document.write( "Please help me </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #371 by </B><A HREF=/tutors/aboutme.mpl?userid=khwang><B>khwang(438)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=khwang><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=371\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  Given ln(3x+8)=ln(2x+2)+ln(x-2)<BR>\n" );
document.write( " Since the right hand side ln(2x+2)+ln(x-2) = ln(2x+2)(x-2) = ln(2x^2 -2x -4),<BR>\n" );
document.write( " So,ln(3x+8) = ln(2x^2 -2x -4) and we have\r<BR>\n" );
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document.write( " 3x+8 = 2x^2 -2x -4, or 2x^2 -5x -12  = 0.<BR>\n" );
document.write( " Factor: (2x + 3)(x - 4) = 0.<BR>\n" );
document.write( " So, x = -3/2 or 4 \r<BR>\n" );
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document.write( " But by the definition of ln (log), 3x +8 , 2(x+1) and x-2 must be positive.<BR>\n" );
document.write( " Hence, x > -8/3, x > -1 and x > 2. Therefore, x must be greater than 2 and<BR>\n" );
document.write( " the solution -3/2 is invalid. The only solution is x =4.\r<BR>\n" );
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document.write( " Check,when x = 4, ln(3x+8)=ln(2x+2)+ln(x-2) becomes<BR>\n" );
document.write( " ln 20 = ln 10 + ln 2 = ln 10*2 ,so 4 is a correct solution of the<BR>\n" );
document.write( " given equation.<BR>\n" );
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