document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1196893>Question 1196893</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The radioactive isotope carbon 14 used to date fossils decays with an annual rate of about 0.000124. If a fossil is found which originally has 2 mg of carbon 14, and it now has 0.18 mg, how old is it? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #829930 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=829930\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "a = initial value<BR>\n" );
document.write( "b = determines if we have growth or decay depending if b > 1 or 0 < b < 1.\r<BR>\n" );
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document.write( "x = number of years<BR>\n" );
document.write( "y = amount of carbon-14 in mg\r<BR>\n" );
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document.write( "Given info:<BR>\n" );
document.write( "a = 2<BR>\n" );
document.write( "b = 1 + r = 1 - 0.000124 = 0.999876<BR>\n" );
document.write( "y = 0.18\r<BR>\n" );
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document.write( "y = a*b^x<BR>\n" );
document.write( "0.18 = 2*0.999876^x<BR>\n" );
document.write( "2*0.999876^x = 0.18<BR>\n" );
document.write( "0.999876^x = 0.18/2<BR>\n" );
document.write( "0.999876^x = 0.09<BR>\n" );
document.write( "log(0.999876^x) = log(0.09)<BR>\n" );
document.write( "x*log(0.999876) = log(0.09)<BR>\n" );
document.write( "x = log(0.09)/log(0.999876)<BR>\n" );
document.write( "x = 19,417.7122011154\r<BR>\n" );
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document.write( "Answer: Approximately 19,417 years old\r<BR>\n" );
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document.write( "Here's an approach using the half-life formula\r<BR>\n" );
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document.write( "The half-life of carbon 14 is approximately H = 5730 years according to these sources here<BR>\n" );
document.write( "<a  rel=nofollow HREF= \"https://pubchem.ncbi.nlm.nih.gov/compound/carbon-14\">https://pubchem.ncbi.nlm.nih.gov/compound/carbon-14</a><BR>\n" );
document.write( "<a  rel=nofollow HREF= \"https://en.wikipedia.org/wiki/Carbon-14\">https://en.wikipedia.org/wiki/Carbon-14</a><BR>\n" );
document.write( "Every 5730 years or so, the amount cuts in half.\r<BR>\n" );
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document.write( "y = a*0.5^(x/H)<BR>\n" );
document.write( "0.18 = 2*0.5^(x/5730)<BR>\n" );
document.write( "0.5^(x/5730) = 0.18/2<BR>\n" );
document.write( "0.5^(x/5730) = 0.09<BR>\n" );
document.write( "log( 0.5^(x/5730) ) = log(0.09)<BR>\n" );
document.write( "(x/5730)*log(0.5) = log(0.09)<BR>\n" );
document.write( "x/5730 = log(0.09)/log(0.5)<BR>\n" );
document.write( "x = 5730*log(0.09)/log(0.5)<BR>\n" );
document.write( "x = 19,905.6257091448<BR>\n" );
document.write( "x = 19,906<BR>\n" );
document.write( "This isn't too far from the 19,417 value calculated earlier. \r<BR>\n" );
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document.write( "On the scale of tens of thousands of years, a few hundred years isn't that much of a difference <BR>\n" );
document.write( "19906-19417 = 488<BR>\n" );
document.write( "489/19906 = 0.0246 = 2.46% error approximately<BR>\n" );
document.write( "Though of course the level of precision will depend on what context you're in. If you're casually talking to a friend, then you don't need that much precision. For scientific papers, then you'll definitely need more accuracy.\r<BR>\n" );
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document.write( "Here's a calculator to help check your work<BR>\n" );
document.write( "<a  rel=nofollow HREF= \"https://www.omnicalculator.com/chemistry/carbon-dating\">https://www.omnicalculator.com/chemistry/carbon-dating</a><BR>\n" );
document.write( "In this case there's 0.18/2 = 0.09 = 9% of the carbon 14 left<BR>\n" );
document.write( "The calculator will produce the result of 19,906<BR>\n" );
document.write( "This result is approximate due to the fact that the half-life 5730 was approximate.<BR>\n" );
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