document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1113287>Question 1113287</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find out how long it takes a &#8203;$2900 investment to earn &#8203;$500 interest if it is invested at 8&#8203;% compounded monthly. Round to the nearest tenth of a year.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #728315 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=728315\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> f = p * (1 + r) ^ n\r<BR>\n" );
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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( "to find the interest earned, you would then subtract the present value from the future value.\r<BR>\n" );
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document.write( "the formula for that would be:\r<BR>\n" );
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document.write( "i = f - p\r<BR>\n" );
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document.write( "solve this equation for f to get:\r<BR>\n" );
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document.write( "f = p + i\r<BR>\n" );
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document.write( "in your problem:\r<BR>\n" );
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document.write( "p = 2900<BR>\n" );
document.write( "f = p + i = 2900 + 500 = 3400<BR>\n" );
document.write( "i = .08 / 12 per month.<BR>\n" );
document.write( "n = number of months.\r<BR>\n" );
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document.write( "your equation becomes:\r<BR>\n" );
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document.write( "3400 = 2900 * (1 + .08/12) ^ n\r<BR>\n" );
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document.write( "divide both sides of this equation by 2900 to get:\r<BR>\n" );
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document.write( "3400 / 2900 = (1 + .08/12) ^ n\r<BR>\n" );
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document.write( "take the log of both sides of this eqution to get:\r<BR>\n" );
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document.write( "log(3400 / 2900) = log((1 +.08/12) ^ n)\r<BR>\n" );
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document.write( "since log (b^x) = x * log(b), this equation becomes:\r<BR>\n" );
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document.write( "log(3400 / 2900) = n * log(1 + .08/12)\r<BR>\n" );
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document.write( "solve for n to get:\r<BR>\n" );
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document.write( "n = log(3400 / 2900) / log(1 + .08/12) = 23.93914847 months.\r<BR>\n" );
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document.write( "replace n in the original equation with that to confirm the solution is correct.\r<BR>\n" );
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document.write( "original equation becomes 3400 = 2900 * (1 + .08/12) ^ 23.93914847.\r<BR>\n" );
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document.write( "this results in 3400 = 3400, confirming the solution is correct.\r<BR>\n" );
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document.write( "23.93914847 months / 12 = 1.994929039 years.\r<BR>\n" );
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document.write( "round this to the nearest tenth of a year to get 2 years.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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