document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=710592>Question 710592</A>: <I><SPAN STYLE=\"word-break: break-all;\"> express\r<BR>\n" );
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document.write( "log27 - log1/9 + (log51 - log17 over log81 - 2log3) + 4log1/3 + 3log3\r<BR>\n" );
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document.write( "in its simplest form </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #437072 by </B><A HREF=/tutors/aboutme.mpl?userid=srinivas.g><B>srinivas.g(540)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=srinivas.g><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=srinivas.g><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=437072\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> log 3^3 -log 3^-2 + (log 51 -log 17  over log 81- log 3^2)+ 4 log 3^-1 + log 3^3<BR>\n" );
document.write( "=log (3^3/3^-2) + (log 51/17 over log (81/9)) + log 3^-4 +log 3^3<BR>\n" );
document.write( "= log 3^5 + log (3) over 2 log 3 )+ log (3^-4 *3^3)<BR>\n" );
document.write( " = log 3^5 + log 3 /2 log 3  + log (3^(-4+3))<BR>\n" );
document.write( "=log 3^5 + 1/2 +log 3^-1<BR>\n" );
document.write( "= 5 log 3 +1/2 -log 3<BR>\n" );
document.write( "=4log 3 +0.5 </SPAN>\n" );
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