document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=250489>Question 250489</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Use the formula N = Ie^(kt), where N is the number of items in terms of the initial population I, at time t, and k is the growth constant equal to the percent of growth per unit of time. A certain radioactive isotope decays at a rate of 0.275% annually. Determine the half-life of this isotope, to the nearest year.<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 #182362 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=182362\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> N = Ie^(kt)\r<BR>\n" );
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document.write( "Let I = 1\r<BR>\n" );
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document.write( "If in 1 year it decays by .275%, then at the end of the first year it will be reduced from 1to 1 - .00275 = .99725\r<BR>\n" );
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document.write( "your formula becomes:\r<BR>\n" );
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document.write( ".99725 = 1 * e^(k*1) which becomes:\r<BR>\n" );
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document.write( ".99725 = 1 * e^k\r<BR>\n" );
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document.write( "which is the same as:\r<BR>\n" );
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document.write( ".99725 = e^k\r<BR>\n" );
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document.write( "take the log of both sides to get:\r<BR>\n" );
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document.write( "log(.99725) = log(e^k)\r<BR>\n" );
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document.write( "this becomes:\r<BR>\n" );
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document.write( "log(.99725) = k*log(e)\r<BR>\n" );
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document.write( "divide both sides by log(e) to get:\r<BR>\n" );
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document.write( "log(.99725) / log(e) = k\r<BR>\n" );
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document.write( "solve for k to get:\r<BR>\n" );
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document.write( "k = -.002753788\r<BR>\n" );
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document.write( "now that you have found k, you can find the half life.\r<BR>\n" );
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document.write( "to find the half life, let I = 1 and let N = .5 and substitute in general equation of N = I * e^(kt) to get:\r<BR>\n" );
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document.write( ".5 = 1 * e^(-.002753788*t)\r<BR>\n" );
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document.write( "this is the same as:\r<BR>\n" );
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document.write( ".5 = e^(-.002753788*t)\r<BR>\n" );
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document.write( "take the log of both sides to get:\r<BR>\n" );
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document.write( "log(.5) = log(e^(-.002753788*t)\r<BR>\n" );
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document.write( "this becomes:\r<BR>\n" );
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document.write( "log(.5) = -.002753788*t*log(e)\r<BR>\n" );
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document.write( "divide both sides of this equation by -.002753788*log(e) to get:\r<BR>\n" );
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document.write( "t = log(.5)/(-.002753788*log(e))\r<BR>\n" );
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document.write( "solve for t to get:\r<BR>\n" );
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document.write( "t = 251.7067876 years\r<BR>\n" );
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document.write( "The half life of this radioactive isotope is 251.70657867 years.\r<BR>\n" );
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document.write( "plug that value in the original equation to get:\r<BR>\n" );
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document.write( "N = 1 * e^(kt) becomes\r<BR>\n" );
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document.write( "N = 1 * e^(-.002753788*251.7067876)\r<BR>\n" );
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document.write( "Solve for N to get:\r<BR>\n" );
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document.write( "N = .5\r<BR>\n" );
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document.write( ".5 is one half of 1 so the half life of the isotope is equal to 251.7067876 years.\r<BR>\n" );
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