document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1040435>Question 1040435</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I'm having wrapping my head around this one and could use your guidance please!\r<BR>\n" );
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document.write( "Suppose $5,400 is invested in an account at an annual interest rate of 3.3% compounded continuously. How long (to the nearest tenth of a year) will it take the investment to double in size? \r<BR>\n" );
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document.write( "Any help is greatly appreciated!\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #655239 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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=655239\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font size=3 color=black><BR>\n" );
document.write( "P = Initial amount deposited = 5400<BR>\n" );
document.write( "A = final amount after t years = 2*5400 = 10800<BR>\n" );
document.write( "r = interest rate in decimal form = 0.033<BR>\n" );
document.write( "t = time in years = unknown\r<BR>\n" );
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document.write( "We're using this formula\r<BR>\n" );
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document.write( "where 'e' is a constant (e = 2.718... approx. It's similar to pi = 3.14...)\r<BR>\n" );
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document.write( "Let's use the values given above to find t\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=10800+=+5400%2Ae%5E%280.033%2At%29 ALIGN=MIDDLE  ALT=\"10800+=+5400%2Ae%5E%280.033%2At%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=10800%2F5400+=+%285400%2Ae%5E%280.033%2At%29%29%2F5400 ALIGN=MIDDLE  ALT=\"10800%2F5400+=+%285400%2Ae%5E%280.033%2At%29%29%2F5400\"   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=0.033t+=+ln%282%29 ALIGN=MIDDLE  ALT=\"0.033t+=+ln%282%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=t+=+ln%282%29%2F0.033 ALIGN=MIDDLE  ALT=\"t+=+ln%282%29%2F0.033\"   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=t+=+21.004460016968 ALIGN=MIDDLE  ALT=\"t+=+21.004460016968\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "It will take approximately 21.004460016968 years.\r<BR>\n" );
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document.write( "Round that to the nearest tenth to get the final answer of <font color=red size=5>21.0 years</font>\r<BR>\n" );
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document.write( "Side note: Using the rule of 72, we'd have 72/x = 72/3.3 = 21.8181818181819 which is fairly close to what we got above. So this rule will give you a quick idea of the approximate doubling time. <BR>\n" );
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