document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=974072>Question 974072</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 2(log3)(logx) = log2 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #596089 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=596089\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 2(log3)(logx) = log2\r<BR>\n" );
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document.write( "log 9=log 3^2=2 log 3\r<BR>\n" );
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document.write( "log 9* log x=log2\r<BR>\n" );
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document.write( "log x=log2/log 9\r<BR>\n" );
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document.write( "log x= 0.301/0.954 = 0.316\r<BR>\n" );
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document.write( "raise everything to the 10 power.  10^log10=1  e^lne=1.\r<BR>\n" );
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document.write( "x= 2.07\r<BR>\n" );
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document.write( "2(0.477)(.316)=0.301\r<BR>\n" );
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document.write( "From years ago, log of 3 is almost one-half, so 2 log 3 (log 9) is almost 1<BR>\n" );
document.write( "Therefore, log x is very nearly log 2, \r<BR>\n" );
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document.write( " </SPAN>\n" );
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