document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=644477>Question 644477</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 7000 is compounded continuously at 6% interest, what is the amount after 2 years? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #405059 by </B><A HREF=/tutors/aboutme.mpl?userid=DrBeeee><B>DrBeeee(684)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=DrBeeee><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=405059\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I think that compunding continuously at p percent per year (expressed as a decimal) has the factor <BR>\n" );
document.write( "(1) e^p, where e = 2.71828... an irrational number named after Euler, a famous Swiss mathematician, 1707-1783. <BR>\n" );
document.write( "The balance, B, of account starting with principal, P, earning p percent per year for n years is<BR>\n" );
document.write( "(2) B = P*(e^p)^n<BR>\n" );
document.write( "For this example we have<BR>\n" );
document.write( "(3) B = 7000*(e^.06)^2<BR>\n" );
document.write( "Using a calculator with the e^x key we get<BR>\n" );
document.write( "(4) B = 7892.478 <BR>\n" );
document.write( "The answer to the nearest cent is $7892.48.<BR>\n" );
document.write( "If you did not compound continuously but only yearly you would have<BR>\n" );
document.write( "(5) By = 7000*(1.06)^2 or<BR>\n" );
document.write( "(6) By = 7865.20<BR>\n" );
document.write( "Giving $27.28 less.<BR>\n" );
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