document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=898387>Question 898387</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve the equation. Use a calculator to obtain decimal approximations, correct to two decimal places.\r<BR>\n" );
document.write( "\n" );
document.write( "ln (2x+3)+ln(x-6)-2lnx = 0\r<BR>\n" );
document.write( "\n" );
document.write( "I know the answer is 10.68 but im just not sure how my teacher got it. Any help would be appreciated. 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 #544811 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=544811\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> your equation is:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "ln(2x+3) + ln(x-6) - 2*ln(x) = 0\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "since ln(2x+3) + ln(x-6) = ln((2x+3) * (x-6)), your equation becomes:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "ln((2x+3)*(x-6)) - 2*ln(x) = 0\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "since 2*ln(x) = ln(x^2) and since ln((2x+3)*(x-6)) - ln(x^2) = ln((2x+3)*(x-6)/x^2), your equation becomes:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "ln((2x+3)*(x-6)/x^2) = 0\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "this is true if and only if:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "e^0 = (2x+3)*(x-6) / x^2\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "since e^0 = 1, your equation becomes:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "1 = (2x+3)*(x-6) / x^2\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "multiply both sides of this equation by x^2 and simplify to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "x^2 = 2x^2 - 12x + 3x - 18 = 2x^2 - 9x - 18\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "subtract x^2 from both sides of this equation to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "0 = x^2 - 9x - 18\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "factor this equation using the quadratic formula to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "x = 10.68465... or x = -1.68465...\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "since x has to be positive, the solution is x = 10.68465... which rounds to 10.68.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );