document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=955318>Question 955318</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please sort these out?(which has the most value , which has the lowest value)<BR>\n" );
document.write( "*log2(3)*          *log3(5)*         *log5(8)*\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #583529 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=583529\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the easiest way to sort this out is to use the log base conversion formula of:\r<BR>\n" );
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document.write( "logb(a) = log10(a)/log10(b) = LOG(a)/LOG(b), where LOG is the log function of your calculator.\r<BR>\n" );
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document.write( "using this formula, you get:\r<BR>\n" );
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document.write( "log2(3) = LOG(3)/LOG(2) = 1.58496...<BR>\n" );
document.write( "log3(5) = LOG(5)/LOG(3) = 1.46973...<BR>\n" );
document.write( "log5(8) = LOG(8)/LOG(5) = 1.29202...\r<BR>\n" );
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document.write( "the one that has the lowest value is the last one.\r<BR>\n" );
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document.write( "you can double check your figures by using the exponential form of the logarithmic equation.\r<BR>\n" );
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document.write( "log2(3) = y if and only if 3 = 2^y<BR>\n" );
document.write( "when y = 1.58496..., 2^y = 3, so we're good there.\r<BR>\n" );
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document.write( "log3(5) = y if and only if 5 = 3^y<BR>\n" );
document.write( "when y = 1.46973..., 3^y = 5, so we're good there.\r<BR>\n" );
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document.write( "log5(8) = y if and only if 8 = 5^y<BR>\n" );
document.write( "when y = 1.29202..., 5^y = 8, so we're good there.\r<BR>\n" );
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document.write( "log5(8) = 1.29202 gives you the smallest value of y.\r<BR>\n" );
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