document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1172513>Question 1172513</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If the number of soap, N, in two baths are modeled by N_a=3000e_{^{0.09t}}\:\:N_b=5000e^{kt}, where 't' is the number from midday. find the value of 'k' for the second bath that will have the number of soap in the two baths being equal at 7:00pm.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #797570 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=797570\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 3000e^(0.09t)=5000e^(kt); t=7<BR>\n" );
document.write( "e^(0.63)=(5/3)e^(7k)<BR>\n" );
document.write( "e^(0.63-7k)=(5/3)<BR>\n" );
document.write( "ln both sides<BR>\n" );
document.write( "0.63-7k=0.5108<BR>\n" );
document.write( "7k=0.1192<BR>\n" );
document.write( "k=0.0170<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28300%2C300%2C-5%2C10%2C-1000%2C10000%2C3000e%5E%280.09x%29%2C5000e%5E%280.017%2Ax%29%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" > </SPAN>\n" );
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