document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=592716>Question 592716</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Condense the expression<BR>\n" );
document.write( "Ln 20+2ln 1/2 - ln x </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #376027 by </B><A HREF=/tutors/aboutme.mpl?userid=mamiya><B>mamiya(56)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mamiya><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=376027\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Condense the expression<BR>\n" );
document.write( "Ln 20+2ln 1/2 - ln x\r<BR>\n" );
document.write( "\n" );
document.write( "  first, we have to remember that exponential logarithm turns a division into a subtraction and  a product into an addition and an exponent into a product, and vice-verca.\r<BR>\n" );
document.write( "\n" );
document.write( "  with that in mind,  we have ,\r<BR>\n" );
document.write( "\n" );
document.write( "  ln 20  +  2ln1/2  -  ln x  =   ln 20 - ln x  +  ln(1/2)^2<BR>\n" );
document.write( "                             =  ln 20/x  +  ln 1/4<BR>\n" );
document.write( "                             = ln (20/x)(1/4)<BR>\n" );
document.write( "                             =  ln (20/4x)<BR>\n" );
document.write( "                             = ln 5/x\r<BR>\n" );
document.write( "\n" );
document.write( "so the condense form of  ln 20 + 2ln1/2 - ln x  is   ln 5/x   </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );