document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209221>Question 1209221</A>: <I><SPAN STYLE=\"word-break: break-all;\"> $3000 is deposited at 8% compounded semiannually. How long does it take to double? Round to the nearest tenth of a year. <BR>\n" );
document.write( "Use the formula<BR>\n" );
document.write( "A=P(1+r/n)^nt<BR>\n" );
document.write( "possible answers<BR>\n" );
document.write( "1.) 8.8 years<BR>\n" );
document.write( "2.) 9.2 years<BR>\n" );
document.write( "3.) 9 years<BR>\n" );
document.write( "4.) 8.6 years </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #847861 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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=847861\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "Answer: <font color=red>8.8 years</font>\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Work Shown<BR>\n" );
document.write( "P = 3000 is the investment amount. We wish to double it to A = 6000 dollars.<BR>\n" );
document.write( "r = 0.08 = decimal form of the annual interest rate<BR>\n" );
document.write( "n = 2 since we're compounding 2 times a year<BR>\n" );
document.write( "The goal is to solve for t.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "A = P(1+r/n)^(nt)<BR>\n" );
document.write( "6000 = 3000(1+0.08/2)^(2*t)<BR>\n" );
document.write( "6000/3000 = (1.04)^(2*t)<BR>\n" );
document.write( "2 = (1.04)^(2*t)<BR>\n" );
document.write( "Log(2) = Log( (1.04)^(2*t) )<BR>\n" );
document.write( "Log(2) = 2t*Log(1.04)<BR>\n" );
document.write( "t = (1/2)*Log(2)/Log(1.04)<BR>\n" );
document.write( "t = 8.836494 approximately<BR>\n" );
document.write( "t = <font color=red>8.8 years</font> when rounding to the nearest tenth of a year. \r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Footnotes:<ul><li>Use logs to isolate the variable exponent. A useful phrase to remember is \"When the variable is in the trees we log it down\".</li><li>I used the rule Log(x^y) = y*Log(x) which allows us to pull down the exponent. </li><li>The logs can be any valid base. Perhaps base 10 is easiest to work with.</li><li>The 3000 doesn't affect the answer. We can change the 3000 to any positive number we want, and get the same result at the end.</li><li>Upon thinking about it more, it doesn't make much sense to round to the nearest tenth. It would be best to round to the nearest half (since we're compounding the money every half a year). However, I'll stick with the answer 8.8 since that appears to be what the teacher wants.</ul><BR>\n" );
document.write( "</font> </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );