document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1166083>Question 1166083</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Log(10+9x)-log(11-x)=2 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #790553 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><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=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=790553\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I'm assuming log here means log base 10<BR>\n" );
document.write( "Using the rule log(a) - log(b) = log(a/b) we have:<BR>\n" );
document.write( "log((10+9x)/(11-x)) = 2<BR>\n" );
document.write( "To eliminate the logarithm, we use the exponential form:<BR>\n" );
document.write( "10^(log((10+9x)/(11-x))) = 10^2<BR>\n" );
document.write( "(10+9x)/(11-x) = 100<BR>\n" );
document.write( "Solving for x gives 1100 - 100x = 10 + 9x -> 109x = 1090 -> x = 10<BR>\n" );
document.write( "Check: log(10+90) - log(1) = 2 </SPAN>\n" );
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