document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1187575>Question 1187575</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A radioactive form of uranium has a half life of 2.5x 10^5 years. a) Find the remaining mass of 1 g sample after t years<BR>\n" );
document.write( "b) Determine the remaining mass of this sample after 5000 years. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #818590 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=818590\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the radioactive form of uranium has a half life of 2.5 * 10^5 years.\r<BR>\n" );
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document.write( "the formula is f = p * e^(rt).\r<BR>\n" );
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document.write( "r is the rate of growth per year.<BR>\n" );
document.write( "t is the time in years.\r<BR>\n" );
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document.write( "if the future value of the uranium is equal to half the present value of the uranium, then the formula becomes:\r<BR>\n" );
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document.write( "1/2 = e^(rt)\r<BR>\n" );
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document.write( "if the time to half life is equal to 2.5 * 10^5 years, then the formula becomes:\r<BR>\n" );
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document.write( "1/2 = e^(r * 2.5 * 10^5)\r<BR>\n" );
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document.write( "take the natural log of both sides of the equation to get:\r<BR>\n" );
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document.write( "ln(1/2) = ln(e ^ (r * 2.5 * 10^5))\r<BR>\n" );
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document.write( "since ln(e ^ (r * 2.5 * 10^5)) is equal to r * 2.5 * 10^5 * ln(e) and since ln(e) = 1, the formula becomes:\r<BR>\n" );
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document.write( "ln(1/2) = r * 2.5 * 10^5\r<BR>\n" );
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document.write( "divide both sides of the equation by (2.5 * 10^5) to get:\r<BR>\n" );
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document.write( "ln(1/2) / (2.5 * 10^5) = r\r<BR>\n" );
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document.write( "solve for r to get:\r<BR>\n" );
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document.write( "r = -2.77258872 * 10^-6\r<BR>\n" );
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document.write( "to see if that value is good, replace r in the original equation with that and find f.\r<BR>\n" );
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document.write( "the original equation is f = p * e^(rt)\r<BR>\n" );
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document.write( "when p = 1, the equation becomes:\r<BR>\n" );
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document.write( "f = e^(rt)\r<BR>\n" );
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document.write( "when r = -2.77258872 * 10^-6 and t = 2.5 * 10^5, the formula becomes:\r<BR>\n" );
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document.write( "f = e ^ (-2.77258872 * 10^-6 * 2.5 * 10^5)\r<BR>\n" );
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document.write( "solve for f to get:\r<BR>\n" );
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document.write( "f = .5.\r<BR>\n" );
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document.write( "the formula is good and the rate of growth is -2.77258872 * 10^-6 per year.\r<BR>\n" );
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document.write( "the remaining mass of 1 gram of uranium after t years is given by the formula:\r<BR>\n" );
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document.write( "f = 1 * e ^ ((-2.77258872 * 10^-6 * t)\r<BR>\n" );
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document.write( "when t = 5000, the formula becomes:\r<BR>\n" );
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document.write( "f = 1 * e ^ ((-2.77258872 * 10^-6 * 5000)\r<BR>\n" );
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document.write( "solve for f to get:\r<BR>\n" );
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document.write( "f = .9862327045 grams.\r<BR>\n" );
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document.write( "you can graph the equation.\r<BR>\n" );
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document.write( "here's what it looks like.<BR>\n" );
document.write( "it shows the values of f (represented by y in the graph) for t (represented by x in the graph) at 5000 years and at 2.5 * 10^5 years in the future.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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