document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1131793>Question 1131793</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Use the formula N=Ie^kt, where N is the number of items of the initial population \"I\", at the time \"t\", and \"k\" is the growth constant equal to the percent of growth per unit of time. A certain radioactive isotope has a half-life of approximately 1750 years. How many years would be required for a given amount of this isotope to decay to 25% of that amount? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #748467 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=748467\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This would be 2 half-lives or 3500 years<BR>\n" );
document.write( "N=Ie^(1750k)<BR>\n" );
document.write( "divide by I and N/I=1/4<BR>\n" );
document.write( "1/2=e^(1750k)<BR>\n" );
document.write( "ln of both sides<BR>\n" );
document.write( "-ln2=1750k<BR>\n" );
document.write( "k=-ln2/1750=-0.000396\r<BR>\n" );
document.write( "\n" );
document.write( "N/I=0.25=e^(0.000396*t)<BR>\n" );
document.write( "ln(0.25)=0.000396t<BR>\n" );
document.write( "-ln4/-0.000396=t=3500.74 years, but 3500 is exact.<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );