document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=530282>Question 530282</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the integer k, k > 2, for, which log (k - 2)! + log(k - 1)! + 2 = 2 log k!. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #350009 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=350009\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I presume that the factorial is inside the log, otherwise it would not easily be defined.\r<BR>\n" );
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document.write( "We can write the original equation,\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \log((k-2)!) + \log((k-1)!) + 2 = 2\log(k!)\">, as\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \log((k-2)!(k-1)!) + 2 = \log((k!)^2)\">, using logarithmic properties. \r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \log((k!)^2) - \log((k-2)!(k-1)!) = 2\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \log(\frac{(k!)^2}{(k-2)!(k-1)!}) = 2\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \log(k(k)(k-1)) = 2\">, upon simplifying the factorial expression in the LHS. Assuming the log is in base 10,\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE k(k)(k-1) = 100\">, in which k = 5 is the unique positive integer solution. </SPAN>\n" );
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