document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=565317>Question 565317</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do I solve:<BR>\n" );
document.write( "Log x + log (13 -3x) = 1\r<BR>\n" );
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document.write( "(Bases are 10) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #365861 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=365861\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> adding logs means multiplying their arguments ___ and ___ 1 = log(10)\r<BR>\n" );
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document.write( "x(13 - 3x) = 10 ___ 13x - 3x^2 = 10 ___ 0 = 3x^2 - 13x + 10\r<BR>\n" );
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document.write( "factoring ___ 0 = (3x - 10)(x - 1) </SPAN>\n" );
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