document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=584722>Question 584722</A>: <I><SPAN STYLE=\"word-break: break-all;\"> find the value(s) of x if 2log3x = log3 (x+6) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #372854 by </B><A HREF=/tutors/aboutme.mpl?userid=Schaman_Dempster><B>Schaman_Dempster(26)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Schaman_Dempster><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=Schaman_Dempster><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=372854\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 2log3x = log3 (x+6)\r<BR>\n" );
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document.write( "Using the property of logarithm such that  n log a  = log  a^n\r<BR>\n" );
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document.write( "log3 x^2 = log3 (x+6)\r<BR>\n" );
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document.write( "Since the logs on both the sides have same bases, they can be canceled out.\r<BR>\n" );
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document.write( "So,  x^2 = x + 6<BR>\n" );
document.write( "   x^2 - x - 6 =0\r<BR>\n" );
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document.write( "Now, factoring by regrouping:<BR>\n" );
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document.write( "     x^2 - 3x + 2x -6 = 0<BR>\n" );
document.write( "  x(x -3) + 2(x-3) = 0<BR>\n" );
document.write( "(x-3) (x +2) = 0<BR>\n" );
document.write( "x = 3 , x =-2\r<BR>\n" );
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document.write( "Since originally we had the term 2 log3 x..which shows that x should always be greater than 0 {Because log of negative number is not defined}\r<BR>\n" );
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document.write( "Thus,  x = 3\r<BR>\n" );
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document.write( "Hope this helps~! </SPAN>\n" );
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