document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=603052>Question 603052</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose that $1300  is invested at an interest rate of 4.5%  per year, compounded continuously. After how many years will the initial investment be doubled?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #380562 by </B><A HREF=/tutors/aboutme.mpl?userid=lwsshak3><B>lwsshak3(11628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=lwsshak3><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=380562\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Suppose that $1300 is invested at an interest rate of 4.5% per year, compounded continuously. After how many years will the initial investment be doubled?<BR>\n" );
document.write( "**<BR>\n" );
document.write( "Formula for continuous compounding: A=Pe^rt, P=initial investment, r=annual interest rate, A=amount after t years<BR>\n" );
document.write( "..<BR>\n" );
document.write( "For given problem:<BR>\n" );
document.write( "A/P=e^.045t=2<BR>\n" );
document.write( ".045t*ln(e)=ln(2)<BR>\n" );
document.write( "ln(e)=1 (log of base=1)<BR>\n" );
document.write( ".045t=ln(2)<BR>\n" );
document.write( "t=ln(2)/.045&#8776;15.4 yrs<BR>\n" );
document.write( "ans:<BR>\n" );
document.write( "The initial investment will double in about 15 years </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );