document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=957761>Question 957761</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Either factor or use the quadratic formula to solve the given equation. (Enter your answers as a comma-separated list.)<BR>\n" );
document.write( "(ln x)^2 + 2 ln x = 3<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 #585294 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=585294\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> sometimes it helps to make your expression equal to a variable so the process of factoring doesn't look so messy.\r<BR>\n" );
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document.write( "if you let y = ln(x), then your equation becomes:\r<BR>\n" );
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document.write( "y^2 + 2y = 3\r<BR>\n" );
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document.write( "subtract 3 from both sides to get:\r<BR>\n" );
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document.write( "y^2 + 2y - 3 = 0\r<BR>\n" );
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document.write( "the equation is now in standard form of ay^2 + by + c = 0, where:\r<BR>\n" );
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document.write( "a = 1<BR>\n" );
document.write( "b = 2<BR>\n" );
document.write( "c = -3\r<BR>\n" );
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document.write( "this allows you to use the quadratic formula if you need it.\r<BR>\n" );
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document.write( "in this case you don't need it because the equation can be factored manually.\r<BR>\n" );
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document.write( "this equation can be factored as (y+3) * (y-1) = 0\r<BR>\n" );
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document.write( "set each of these factors to 0 and you get (y+3) = 0 and (y-1) = 0\r<BR>\n" );
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document.write( "solve for y to get y = -3 and y = 1\r<BR>\n" );
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document.write( "replace y with ln(x) that you made it equal to in the beginning.\r<BR>\n" );
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document.write( "y = -3 becomes ln(x) = -3\r<BR>\n" );
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document.write( "y = 1 becomes ln(x) = 1\r<BR>\n" );
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document.write( "solve for x in each of these equations.\r<BR>\n" );
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document.write( "ln(x) = -3 if and only if e^(-3) = x which results in x = .0497870684.\r<BR>\n" );
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document.write( "ln(x) = 1 if and only if e^1 = x which results in x = 2.718281828.\r<BR>\n" );
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document.write( "replace each of these values for x in the original equation to see if those original equations hold true.\r<BR>\n" );
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document.write( "the original equation is ln(x)^2 + 2*ln(x) = 3\r<BR>\n" );
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document.write( "replace x with .0497870684 and you get ln(x)^2 + 2*ln(x) = 3 becomes 3 = 3.\r<BR>\n" );
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document.write( "replace x with 2.718281828 and you get ln(x)^2 + 2*ln(x) = 3 becomes 3 = 3.\r<BR>\n" );
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document.write( "both values of x are good so they are your solutions.\r<BR>\n" );
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