document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1057999>Question 1057999</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose that $2000 is invested in an account that pays interest compounded continuously. Find the amount of time that it would take for the account to grow to $4000 at 3.75 %. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #673014 by </B><A HREF=/tutors/aboutme.mpl?userid=solve_for_x><B>solve_for_x(190)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solve_for_x><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=673014\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Use the equation for continuous compounding:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+Pe%5E%28rt%29 ALIGN=MIDDLE  ALT=\"A+=+Pe%5E%28rt%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "with A = 4000, P = 2000, and r = 0.0375.\r<BR>\n" );
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document.write( "Substituting the value for A, P, and r into the equation and<BR>\n" );
document.write( "solving for t gives:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=4000+=+2000%2Ae%5E%280.0375t%29 ALIGN=MIDDLE  ALT=\"4000+=+2000%2Ae%5E%280.0375t%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2+=+e%5E%280.0375t%29 ALIGN=MIDDLE  ALT=\"2+=+e%5E%280.0375t%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "ln(2) = 0.0375t\r<BR>\n" );
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document.write( "t = ln(2) / 0.0375\r<BR>\n" );
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document.write( "t = 18.48 years </SPAN>\n" );
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