document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1078094>Question 1078094</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do you expand log10 x+log10(3x-5)=log10^2? I tried to solve this equation but I can't figure it out.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #692564 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=692564\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  log10 x+log10(3x-5)=log10 2<BR>\n" );
document.write( "adding logs is multiplication of what the log is of.<BR>\n" );
document.write( "x(3x-5)=2, by raising everything to the 10 power.<BR>\n" );
document.write( "3x^2-5x-2=0<BR>\n" );
document.write( "(3x+1)(x-2)=0<BR>\n" );
document.write( "x=-1/3, 2.  Logs can't be negative, so x=2<BR>\n" );
document.write( "check<BR>\n" );
document.write( "log10 2+ log10(1)=log10 2<BR>\n" );
document.write( "log10 (1)=0<BR>\n" );
document.write( "log10 2=1og 10 2<BR>\n" );
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