document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=194020>Question 194020</A>: <I><SPAN STYLE=\"word-break: break-all;\"> You invest $1,000 at an annual rate of 6.75% compounded continuously.How long will it take for the value of the investment to double?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #145644 by </B><A HREF=/tutors/aboutme.mpl?userid=RAY100><B>RAY100(1637)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=RAY100><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=145644\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> formula  for  continuously  compounding  is  ,,,A = Pe^rt<BR>\n" );
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document.write( "2000 = 1000 e^(.0675*t)<BR>\n" );
document.write( "ln  of  both  sides<BR>\n" );
document.write( "ln2=.0675 *t,,,,,,,,,,,,,,,,,,,,,,,,          lne^.0675t=.0675t<BR>\n" );
document.write( "ln2 / .0675 = t<BR>\n" );
document.write( "10.26  =  t  ,,,,,,,in  yrs<BR>\n" );
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document.write( "check<BR>\n" );
document.write( "2000 =1000 e^(.0675*10.26)=1000 * 2 = 2000,,,,,,ok </SPAN>\n" );
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