document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=20558>Question 20558</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What is M equal to in: log<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6 ALIGN=MIDDLE  ALT=\"6\"   DISABLED_style= \"max-width:90%; height:auto;\" >(3m+7) - log<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6 ALIGN=MIDDLE  ALT=\"6\"   DISABLED_style= \"max-width:90%; height:auto;\" >(m+4) = 2log<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6 ALIGN=MIDDLE  ALT=\"6\"   DISABLED_style= \"max-width:90%; height:auto;\" >6 - 3log<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6 ALIGN=MIDDLE  ALT=\"6\"   DISABLED_style= \"max-width:90%; height:auto;\" >3 I was thinking that you add log<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6 ALIGN=MIDDLE  ALT=\"6\"   DISABLED_style= \"max-width:90%; height:auto;\" >(m+4) to both sides, but i believe that isn't right.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #10198 by </B><A HREF=/tutors/aboutme.mpl?userid=venugopalramana><B>venugopalramana(3286)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=venugopalramana><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.geocities.com/venugopalramana/FREE_MATHS_ASSISTANCE.html valign=middle><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=10198\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> log<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6 ALIGN=MIDDLE  ALT=\"6\"   DISABLED_style= \"max-width:90%; height:auto;\" >(3m+7) - log<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6 ALIGN=MIDDLE  ALT=\"6\"   DISABLED_style= \"max-width:90%; height:auto;\" >(m+4) = 2log<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6 ALIGN=MIDDLE  ALT=\"6\"   DISABLED_style= \"max-width:90%; height:auto;\" >6 - 3log<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6 ALIGN=MIDDLE  ALT=\"6\"   DISABLED_style= \"max-width:90%; height:auto;\" >3 I was thinking that you add log<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6 ALIGN=MIDDLE  ALT=\"6\"   DISABLED_style= \"max-width:90%; height:auto;\" >(m+4) to both sides, but i believe that isn't right. <BR>\n" );
document.write( "use the formula log x to base y =log x/log y where all logs are to any common base.<BR>\n" );
document.write( "so we get {(log (3m+7)/log 6)}-{(log (m+4)/log 6)}={(2 log 6/log 6) -(3 log 3/log 6)}<BR>\n" );
document.write( "now multiply through out with log 6 to get <BR>\n" );
document.write( "log (3m+7)-log (m+4)=2log 6-3log 3<BR>\n" );
document.write( "now use the formula log x^n = n*log x<BR>\n" );
document.write( "and  log(x/y)=log x-log y<BR>\n" );
document.write( "log ((3m+7)/(m+4))=log 6*6-log 3*3*3=log 36-log 27=log(36/27)<BR>\n" );
document.write( "taking antilogs on both sides <BR>\n" );
document.write( "((3m+7)/(m+4))=36/27=4/3<BR>\n" );
document.write( "cross multiplying<BR>\n" );
document.write( "3(3m+7)=4(m+4)<BR>\n" );
document.write( "9m+21=4m+16<BR>\n" );
document.write( "9m-4m=16-21=-5<BR>\n" );
document.write( "5m=-5<BR>\n" );
document.write( "m=-1<BR>\n" );
document.write( " </SPAN>\n" );
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