document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=972610>Question 972610</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Thomas wants to save money for a vacation. Thomas invests $1,200 in an account that pays an interest rate of 4%.<BR>\n" );
document.write( "How many years will it take for the account to reach $14,000? Round your answer to the nearest hundredth.<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 #594896 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=594896\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A=p{1+(r/n)}^nt   where r is rate, n the number of compoundings per year and t the years.\r<BR>\n" );
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document.write( "14000=1200{1+.04/n}^nt    We are not given the number of compoundings.  I will assume 1.\r<BR>\n" );
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document.write( "14000=1200(1.04)^t\r<BR>\n" );
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document.write( "11.6667=1.04^t<BR>\n" );
document.write( "logs of both sides<BR>\n" );
document.write( "log 11.6667=t log 1.04<BR>\n" );
document.write( "log11.6667/log 1.04 =t<BR>\n" );
document.write( "1.067/.017=62.76 years\r<BR>\n" );
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document.write( "with continuous compounding<BR>\n" );
document.write( "A=Pe^rt<BR>\n" );
document.write( "11.6667=e^rt<BR>\n" );
document.write( "ln of both sides<BR>\n" );
document.write( "2.456=.04*t<BR>\n" );
document.write( "t=61.40 years\r<BR>\n" );
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document.write( "Rule of 72, it would double in 18 years.  Three doublings in 54 years, and four doublings in 72 years. <BR>\n" );
document.write( "That would be 9600 in 54 years and 19,200 in 72 years, so the answer is between them, as it is. </SPAN>\n" );
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