document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=237761>Question 237761</A>: <I><SPAN STYLE=\"word-break: break-all;\"> hi i am trying to figure out where to go with this problem ln6+ln(x-2)=ln(3x-2).  it seems like i can get it to ln6=2x but from there it turns out to be x=.8958 but the answer is supposed to come out to be x=10/3 where am i going wrong? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #174807 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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/cgi-bin/show-face.mpl?user=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=174807\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> adding logs is analogous to multiplying quantities\r<BR>\n" );
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document.write( "ln(6 (x - 2)) = ln(3x - 2)\r<BR>\n" );
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document.write( "6x - 12 = 3x - 2\r<BR>\n" );
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document.write( "3x = 10\r<BR>\n" );
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document.write( "x = 10/3 </SPAN>\n" );
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