document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=81933>Question 81933</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Here is the offending problem:\r<BR>\n" );
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document.write( "6e^5t = 18e^3t+2\r<BR>\n" );
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document.write( "I'm not sure how to start this one. My professor didn't give us an example like this that I can reference.  Here is my attempt:\r<BR>\n" );
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document.write( "6e^5t = 18e^3t+2<BR>\n" );
document.write( "6e^5t - 18e^3t+2 =0<BR>\n" );
document.write( "6(e^5t - 2e^3t+2) = 0\r<BR>\n" );
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document.write( "At this point I am lost. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #58715 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=58715\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 6e^(5t)=18e^(3t+2) ... dividing by 6 gives e^(5t)=2e^(3t+2) ... taking natural log gives 5t=ln(2)+3t+2\r<BR>\n" );
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document.write( "subtracting 3t gives 2t=ln(2)+2 ... so t=1+((ln(2))/2) </SPAN>\n" );
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