document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=301769>Question 301769</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve the problem.\r<BR>\n" );
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document.write( "Use the formula N = Iekt, 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" );
document.write( "        182 years <BR>\n" );
document.write( "        3 years <BR>\n" );
document.write( "        252 years <BR>\n" );
document.write( "        109 years <BR>\n" );
document.write( "Thank you<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 #216605 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><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=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=216605\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Use the formula N = Iekt, 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. <BR>\n" );
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document.write( "I think the formula should be N = I*e^(kt)<BR>\n" );
document.write( "Find the growth constant, k using t = 1 yr<BR>\n" );
document.write( "I = 1<BR>\n" );
document.write( "then<BR>\n" );
document.write( "N = 1 - .00275 = .99725 remaining after 1 hr<BR>\n" );
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document.write( "e*k = .99725<BR>\n" );
document.write( "k*ln(e) = ln(.99725)<BR>\n" );
document.write( "k = -.00275, (decay is negative growth)<BR>\n" );
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document.write( "A certain radioactive isotope decays at a rate of 0.275% annually.<BR>\n" );
document.write( " Determine the half-life of this isotope, to the nearest year.<BR>\n" );
document.write( "N = .5<BR>\n" );
document.write( "I = 1<BR>\n" );
document.write( "k = -.00275<BR>\n" );
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document.write( "1*e^(-.00275t) = .5<BR>\n" );
document.write( "-.00275t*ln(e) = ln(.5)<BR>\n" );
document.write( "-.00275t = -.693<BR>\n" );
document.write( "t = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-.693%29%2F%28-.00275%29 ALIGN=MIDDLE  ALT=\"%28-.693%29%2F%28-.00275%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "t = 252.05 yrs is the half life of the isotope<BR>\n" );
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document.write( "Check solution on a calc<BR>\n" );
document.write( "enter e^(-.00275*252) = .500; confirms our solutions\r<BR>\n" );
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