document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1144054>Question 1144054</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A 1000 pound fund is loaned at 5% interest for 100 years.  After 100 yrs., the fund would be worth over 100 times the original amount.  Calculate amt.\r<BR>\n" );
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document.write( "Interest at 5% = .05\r<BR>\n" );
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document.write( "Not sure how to solve.  Non-homework.\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 #765047 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=765047\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the general formula for this type of problem is f = p * (1 + r) ^ n.<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( "n is the number of time periods.\r<BR>\n" );
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document.write( "your time period is in years.<BR>\n" );
document.write( "n is equal to 100 years.<BR>\n" );
document.write( "r is equal to .05 per year.<BR>\n" );
document.write( "p is equal to 1000 pounds.<BR>\n" );
document.write( "you want to find f.\r<BR>\n" );
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document.write( "your formula becomes f = 1000 * (1 + .05) ^ 100.\r<BR>\n" );
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document.write( "solve for f to get f = 131,501.2578\r<BR>\n" );
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document.write( "that is indeed greater than 100,000 which is equal to 1,000 * 100.\r<BR>\n" );
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document.write( "your statement is therefore accurate.\r<BR>\n" );
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document.write( "if you wanted to find out when you actually made 100 times your original amount, you would do the following.\r<BR>\n" );
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document.write( "f = 100,000<BR>\n" );
document.write( "p = 1,000<BR>\n" );
document.write( "r = .05<BR>\n" );
document.write( "n = what you want to find.\r<BR>\n" );
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document.write( "the same formula is used.<BR>\n" );
document.write( "it becomes 100,000 = 1,000 * (1 + .05) ^ n<BR>\n" );
document.write( "you want to find n.<BR>\n" );
document.write( "you would divide both sides of this equation by 100 to get:<BR>\n" );
document.write( "100,000 / 1,000 = (1 + .05) ^ n<BR>\n" );
document.write( "simplify to get 100 = (1 + .05) ^ n<BR>\n" );
document.write( "take the log of both sides of this equation to get:<BR>\n" );
document.write( "log(100) = log((1 + .05) ^ n)<BR>\n" );
document.write( "since log((1 + .05) ^ n) = n * log(1 + .05), the formula becomes:<BR>\n" );
document.write( "log(100) = n * log(1.05)<BR>\n" );
document.write( "divide both sides of this equation by log(1.05) to get:<BR>\n" );
document.write( "log(100) / log(1.05) = n<BR>\n" );
document.write( "solve for n to get n = 94.38726564\r<BR>\n" );
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document.write( "the future value of your investment should be exactly 100 times the initial investment in 94.38726564 years.\r<BR>\n" );
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document.write( "confirm this is true by replacing n in the original equation with that value to get:<BR>\n" );
document.write( "f = 1,000 * (1 + .05) ^ 94.38726564<BR>\n" );
document.write( "solve for f to get f = 100,000.\r<BR>\n" );
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document.write( "the solution to when you would actually get exactly 100 times your initial investment is confirmed to be good.\r<BR>\n" );
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document.write( "you can also graph the equation to visually see what happens to your investment.<BR>\n" );
document.write( "for graphing purposes, the equation becomes y = 1000 * 1.05 ^ x.<BR>\n" );
document.write( "this is what the graph looks like.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2019/082002.jpg\" alt=\"$$$\" >\r<BR>\n" );
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document.write( "as you can see, your investment really starts taking off after about the 60th year. </SPAN>\n" );
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