document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1159783>Question 1159783</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If you invest $2,500 in an account, what is the time that will take you to get $2,571 if the<BR>\n" );
document.write( "interest rate is 0.7% compounded continuously? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #782856 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=782856\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the continuous compounding formula is f = p * e ^ (r * t)<BR>\n" );
document.write( "f is the future value.<BR>\n" );
document.write( "p is the present value.<BR>\n" );
document.write( "r is the interest rate per time period.<BR>\n" );
document.write( "t is the number of time periods.\r<BR>\n" );
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document.write( "in this problem, the time period is in years and the formula becomes:<BR>\n" );
document.write( "2571 = 2500 * e ^ (.007 * t)\r<BR>\n" );
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document.write( "the formula says that 2500 becomes 2571 in t years at the rate of .007 per year.\r<BR>\n" );
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document.write( "in this formula, you are using the rate rather than the rate percent.<BR>\n" );
document.write( "the rate is equal to the rate percent divided by 100.<BR>\n" );
document.write( "that makes r = .7% / 100 = .007.\r<BR>\n" );
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document.write( "divide both sides of this formula by 2500 to get:<BR>\n" );
document.write( "2571 / 2500 = e ^ (.007 * t)<BR>\n" );
document.write( "take the natural log of both sides of this equation to get:<BR>\n" );
document.write( "ln(2571/2500) = ln(e^(.007*t)).<BR>\n" );
document.write( "since ln(e^(.007*t)) is equal to t * ln(e^(.007)), the formula becomes:<BR>\n" );
document.write( "ln(2571/2500) = t * ln(e^(.007)).<BR>\n" );
document.write( "solve for t to get:<BR>\n" );
document.write( "t = ln(2571/2500) / ln(e^(.007)) = 4.000599487 years.\r<BR>\n" );
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document.write( "to confirm, replace t in the original equation with that to get:<BR>\n" );
document.write( "2571 = 2500 * e^(.007 * 4.000599487) = 2571.<BR>\n" );
document.write( "the solution is confirmed to be good.<BR>\n" );
document.write( "the solution is that 2500 will become 2571 in 4.000599487 years at the rate of .7% per year compounded continuously.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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