document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=134126>Question 134126</A>: <I><SPAN STYLE=\"word-break: break-all;\"> $6,000 is put into a bank at an annual percentage rate of 13%. Find time required for investment to double if interest is compounded continously?????\r<BR>\n" );
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document.write( "I know the formula is something like A=Pe^rt where r is interest rate and t is years...but what should i do with it? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #98096 by </B><A HREF=/tutors/aboutme.mpl?userid=oscargut><B>oscargut(2103)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=oscargut><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=oscargut><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=98096\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 12000=6000e^(0,13t)<BR>\n" );
document.write( "then<BR>\n" );
document.write( "e^(0,13t)=2<BR>\n" );
document.write( "then<BR>\n" );
document.write( "0,13t=L(2)<BR>\n" );
document.write( "t-L(2)/0,13=5,33years (aprox)  </SPAN>\n" );
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