document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=186596>Question 186596</A>: <I><SPAN STYLE=\"word-break: break-all;\"> log3 + 2 = log x\r<BR>\n" );
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document.write( "solve for x\r<BR>\n" );
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document.write( "logs are to base 10 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #139859 by </B><A HREF=/tutors/aboutme.mpl?userid=nerdybill><B>nerdybill(7384)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=nerdybill><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=nerdybill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=139859\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You will need to apply \"log rules\".<BR>\n" );
document.write( "Review them at:<BR>\n" );
document.write( "http://www.purplemath.com/modules/logrules.htm<BR>\n" );
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document.write( "log3 + 2 = log x <BR>\n" );
document.write( "log3 - log x = -2<BR>\n" );
document.write( "log(3/x) = -2<BR>\n" );
document.write( "3/x = 10^(-2)<BR>\n" );
document.write( "3/x = 0.01<BR>\n" );
document.write( "3 = 0.01x<BR>\n" );
document.write( "3/0.01 = x<BR>\n" );
document.write( "300 = x<BR>\n" );
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