document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=468194>Question 468194</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I am struggling with the following problem:\r<BR>\n" );
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document.write( "2e^2x+5e^x-12=0\r<BR>\n" );
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document.write( "I know that we replace e^ with u, then use quadratic formula to solve. We get u=6 or u=-16.\r<BR>\n" );
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document.write( "Then, I guess we scratch the -16 because it's negative.  From there I'm not sure how we are supposed to solve for x. I know that the answer is ln (3/2), but I'm not sure of the steps that go into solving for that and why the answer isn't ln3/2 or ln3/ln2, etc. Obviously you get here by setting 6=2e^2x, but I'm not sure why. Just need a brush-up on some rules here.\r<BR>\n" );
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document.write( "Thanks,\r<BR>\n" );
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document.write( "J </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #321281 by </B><A HREF=/tutors/aboutme.mpl?userid=mananth><B>mananth(16946)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mananth><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=mananth><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=321281\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 2e^2x+5e^x-12=0<BR>\n" );
document.write( "let e^x=u\r<BR>\n" );
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document.write( "2u^2-5u-12=0\r<BR>\n" );
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document.write( "2u^2-8u+3u-12=0<BR>\n" );
document.write( "2u(u-4)+3(u-4)=0\r<BR>\n" );
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document.write( "(u-4)(2u+3)=0<BR>\n" );
document.write( "e^x=4 \r<BR>\n" );
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document.write( "ln(e^x)= ln(4)<BR>\n" );
document.write( "x= ln(4)\r<BR>\n" );
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document.write( "e^x= -3/2<BR>\n" );
document.write( "x=ln(-3/2) not defined </SPAN>\n" );
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