document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1187659>Question 1187659</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The half-life of two products of a chemical disaster are shown in the chart below:<BR>\n" );
document.write( "Substance Half-life<BR>\n" );
document.write( "Iodine-131 8.1 days<BR>\n" );
document.write( "Cesium-144 282 days<BR>\n" );
document.write( "Draw a graph showing the percent remaining during the first 5 half-lives.<BR>\n" );
document.write( "b. Express the percent remaining as a function of the number of half-lives<BR>\n" );
document.write( "elapsed n.<BR>\n" );
document.write( "c. What percent of the substance remains after one week?<BR>\n" );
document.write( "d. How long is it until the percent remaining of each substance is 10%? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #818773 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=818773\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> formula for growth is f = p * (1 + r) ^ n\r<BR>\n" );
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document.write( "in this problem, the time period is in days.\r<BR>\n" );
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document.write( "to find the rate for the half life, make f = 1/2 and p = 1.\r<BR>\n" );
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document.write( "formula becomes 1/2 = 1 * (1 + r) ^ n\r<BR>\n" );
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document.write( "since 1 * anything is anything, the formula becomes:\r<BR>\n" );
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document.write( "1/2 = (1 + r) ^ n\r<BR>\n" );
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document.write( "for iodine, the half life is represented by 1/2 = (1 + r) ^ 8.l\r<BR>\n" );
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document.write( "for cesium, the half life is represented by 1/2 = (1 + r) ^ 282\r<BR>\n" );
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document.write( "when you solve for r, you get the rate of growth per day.\r<BR>\n" );
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document.write( "a rate of growth less than 1 is a decay.\r<BR>\n" );
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document.write( "to find the rate of growth for iodine, raise both sides of the equation to the (1/8.1) power to get:\r<BR>\n" );
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document.write( "(1/2) ^ (1/8.1) = (1 + r)\r<BR>\n" );
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document.write( "solve for (1 + r) to get:\r<BR>\n" );
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document.write( "(1 + r) = .9179854612.\r<BR>\n" );
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document.write( "to find the rate of growth for cesium raise both sides of the equation to the (1/282) power to get:\r<BR>\n" );
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document.write( "(1/2) ^ (1/282) = (1 + r)\r<BR>\n" );
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document.write( "solve for (1 + r) to get:\r<BR>\n" );
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document.write( "(1 + r) = .9975450496.\r<BR>\n" );
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document.write( "iodine will have half its life remaining every 8.1 days.\r<BR>\n" );
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document.write( "cesium will have half its life remaining every 282 days.\r<BR>\n" );
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document.write( "the growth formula for iodine will be f = .9179854612^(8.1*x).\r<BR>\n" );
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document.write( "the growth formula for cesium will be f = .9975450496^(282*x).\r<BR>\n" );
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document.write( "these formulas will give you the percent remaining for each half life up to 5 half lives.\r<BR>\n" );
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document.write( "the graph of these equations are shown below:\r<BR>\n" );
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document.write( "the coordinate points on these graphs tell you the percent remaining based on the number of half lives that have elapsed.\r<BR>\n" );
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document.write( "for example, when x = 1, .5 = 50 percent is remaining and when x = 5, .0312 or 3.12% is remaining.\r<BR>\n" );
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document.write( "these results are rounded in the graph.\r<BR>\n" );
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document.write( "to get a more detailed answer, make a table as shown below.\r<BR>\n" );
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document.write( "x = 1, 50% remaining.<BR>\n" );
document.write( "x = 2, 25% remaining.<BR>\n" );
document.write( "x = 3, 12.5% remaining.<BR>\n" );
document.write( "x = 4, 6.25% remaining.<BR>\n" );
document.write( "x = 5, 3.125% remaining.\r<BR>\n" );
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document.write( "the graph was accurate except for the last one, where it looks like it truncated the 5 from 3.125 to get 3.12, or .0312 as shown on the graph.\r<BR>\n" );
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document.write( "the graph showed the rate.<BR>\n" );
document.write( "the percent is 100 * the rate.\r<BR>\n" );
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document.write( "x represented the number of half lives.<BR>\n" );
document.write( "it was the same in both graphs, except the exponent in the first graph was 8.1 * x and the exponent in the second graph was 282 * x.\r<BR>\n" );
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document.write( "that's because the half life of the iodine was every 8.1 days and the half life of the cesium was every 282 days.\r<BR>\n" );
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document.write( "to find the percent remaining after 7 days, the formulas to use are:\r<BR>\n" );
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document.write( "for iodine, f = .9179854612^x\r<BR>\n" );
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document.write( "for cesium, f = .9975450496^x\r<BR>\n" );
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document.write( "the x represents the number of days.\r<BR>\n" );
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document.write( "the growth rate shown is the growth per day.\r<BR>\n" );
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document.write( "for iodine, the half life will be achieve for every multiple of x = 8.1.\r<BR>\n" );
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document.write( "for cesium, the half life will be achieved for every multiple of x = 282.\r<BR>\n" );
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document.write( "here are two graphs that show that, the first for iodine, the second for cesium.\r<BR>\n" );
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document.write( "the following graphs show the percent remaining after 7 days.\r<BR>\n" );
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document.write( "the first graph is for iodine, where the percent remaining will be f = .9179854612^7 = .5493518875 * 100 = 54.93518875%.\r<BR>\n" );
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document.write( "the second graph is for cesium, where the percent remaining will be f = .9975450496^7 = .9829413931 * 100 = 98.29413931%.\r<BR>\n" );
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document.write( "the results on the graph are the rate, not the percent, and they are rounded to 3 decimal places; .549 for iodine and .983 for cesium.\r<BR>\n" );
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document.write( "to find  how long it takes to have 10% of the original product remaining, the formulas will be:\r<BR>\n" );
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document.write( "for iodine, f = .9179854612^x becomes .10 = .9179854612^x.\r<BR>\n" );
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document.write( "for cesium, f = .9975450496^x becomes .10 = .9975450496^x. \r<BR>\n" );
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document.write( "to find the value of x, you use logs.\r<BR>\n" );
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document.write( "take the log of both sides of the equation to get:\r<BR>\n" );
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document.write( "for iodine:<BR>\n" );
document.write( "log(.10) = log(.9179854612^x). <BR>\n" );
document.write( "this becomes:<BR>\n" );
document.write( "log(.10) = x * log(.9179854612).<BR>\n" );
document.write( "solve for x to get:<BR>\n" );
document.write( "x = log(.10) / log(.9179854612) = 26.90761757.<BR>\n" );
document.write( "that's the number of days to get to 10% of the original amount.\r<BR>\n" );
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document.write( "for cesium:<BR>\n" );
document.write( "log(.10) = log(.9975450496^x).<BR>\n" );
document.write( "this becomes:<BR>\n" );
document.write( "log(.10) = x * log(.9975450496).<BR>\n" );
document.write( "solve for x to get:<BR>\n" );
document.write( "x = log(.10) / log(.9975450496) = 936.7837228.<BR>\n" );
document.write( "that's the number of days to get to 10% of the original amount.\r<BR>\n" );
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document.write( "the following graphs show those results.\r<BR>\n" );
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document.write( "the first is for iodine; the second is for cesium.\r<BR>\n" );
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document.write( "let me know if you have any questions.\r<BR>\n" );
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document.write( "theo\r<BR>\n" );
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document.write( " </SPAN>\n" );
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