document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=452156>Question 452156</A>: <I><SPAN STYLE=\"word-break: break-all;\"> how do i find x? log[13](3x+9)-log[13](3)=log[13](12) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #310945 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=310945\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> log[13](3x+9)-log[13](3)=log[13](12)<BR>\n" );
document.write( "log(3x+9) - log(3) = log(12) All bases are the same, so it doesn't matter.<BR>\n" );
document.write( "Subtracting logs --> division<BR>\n" );
document.write( "log( (3x+ 9)/3) = log(12)<BR>\n" );
document.write( "If the logs are equal, then<BR>\n" );
document.write( "(3x+9)/3 = 12<BR>\n" );
document.write( "x+3 = 12<BR>\n" );
document.write( "x = 9<BR>\n" );
document.write( " </SPAN>\n" );
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