document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1102759>Question 1102759</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 4. $1000 is invested at 0.025% per annum interest, compounded quarterly. <BR>\n" );
document.write( "a. Calculate the value of the investment at the end of 12 years. <BR>\n" );
document.write( "b. Calculate the minimum number of years required for the value of the investment to exceed $10,000. <BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #717506 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=717506\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Compound interest formula is P=Po(1+r/n)^nt<BR>\n" );
document.write( "P=1000(1+(0.00025)/4)^48=$1003.<BR>\n" );
document.write( "Perhaps it is 2.5% interest or 0.025<BR>\n" );
document.write( "Then it is 1000(1+.025/4)^48=$1348.60.<BR>\n" );
document.write( "Using the 2.5% interest<BR>\n" );
document.write( "10000=1000(1+.025/4)^4t, where t is number of years<BR>\n" );
document.write( "10=1.00625^4t<BR>\n" );
document.write( "ln both sides<BR>\n" );
document.write( "2.303=4t*ln(1.00625)<BR>\n" );
document.write( "369.56 compoundings=4t<BR>\n" );
document.write( "t=92.39 years round to 93 years<BR>\n" );
document.write( "rule of 72 says 72/2.5=doubling time in years which is about 28 years.<BR>\n" );
document.write( "doubles a second time at 56 years (to $4000) and again at 84 years (to $8000), so that 93 years is a reasonable answer.<BR>\n" );
document.write( "(for the 0.00025 interest, it would be 9210 years.) </SPAN>\n" );
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