document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=2882>Question 2882</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What is (log3 x)^2 = log3 x^2 + 3.<BR>\n" );
document.write( "The answer in the answers part is 27 and 1/3.<BR>\n" );
document.write( "Thank you. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #1261 by </B><A HREF=/tutors/aboutme.mpl?userid=khwang><B>khwang(438)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=khwang><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=1261\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  Since log3 x^2 = 2 log3 x,<BR>\n" );
document.write( " (log3 x)^2 = log3 x^2 + 3 converts to\r<BR>\n" );
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document.write( " (log3 x)^2 - 2log3 x - 3 = 0.\r<BR>\n" );
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document.write( " Let u = log3 x, we have u^2 - 2u -3 =0,<BR>\n" );
document.write( " Factoring (u-3)(u+1) =0,<BR>\n" );
document.write( " So, u =3 or -1. <BR>\n" );
document.write( " Hence, log3 x = 3,u = 3^3 = 27 or<BR>\n" );
document.write( " log3 x = -1,u = 3^(-1) = 1/3\r<BR>\n" );
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document.write( " </SPAN>\n" );
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