document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=162053>Question 162053</A>: <I><SPAN STYLE=\"word-break: break-all;\"> CAN SOMEONE HELP ME WITH THIS PROBLEM\r<BR>\n" );
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document.write( "In the formula A = I^ekt, A is the amount of radioactive material remaining from an initial amount I at a given time t and k is a negative constant determined by the nature of the material. An artifact is discovered at a certain site. If it has 53% of the carbon-14 it originally contained, what is the approximate age of the artifact? (carbon-14 decays at the rate of 0.0125% annually.) (Round to the nearest year.) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #119651 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=119651\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> In the formula A = I^ekt, A is the amount of radioactive material remaining from an initial amount I at a given time t and k is a negative constant determined by the nature of the material. An artifact is discovered at a certain site. If it has 53% of the carbon-14 it originally contained, what is the approximate age of the artifact? (carbon-14 decays at the rate of 0.0125% annually.) (Round to the nearest year.)<BR>\n" );
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document.write( "I think the formula should be A = I*e^(-kt)<BR>\n" );
document.write( "In this problem:<BR>\n" );
document.write( "Assume I = 1<BR>\n" );
document.write( "A = .53<BR>\n" );
document.write( "Decimal value for .0125% = .000125<BR>\n" );
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document.write( "e^(-.000125t) = .53<BR>\n" );
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document.write( "-.000125t*ln(e) = ln(.53)<BR>\n" );
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document.write( "-.000125t = -.6834878; (the ln of e is 1)<BR>\n" );
document.write( "t = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-.6834878%29%2F%28-.000125%29 ALIGN=MIDDLE  ALT=\"%28-.6834878%29%2F%28-.000125%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "t = 5,079 years<BR>\n" );
document.write( " </SPAN>\n" );
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