document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=64487>Question 64487</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0495012661/algebrahomeworkh>college algebra</A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> I have used this sight before but it has been awhile.  I having trouble with this carbon-14 dating question.  Can anyone help me.\r<BR>\n" );
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document.write( "We are using 5715 has a half life of carbon-14.  \r<BR>\n" );
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document.write( "A mummy was found with only 24.6% of carbon-14.  How long ago did the person die, to the nearest year. \r<BR>\n" );
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document.write( "Thanks </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #45121 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=45121\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> We are using 5715 has a half life of carbon-14. <BR>\n" );
document.write( "A mummy was found with only 24.6% of carbon-14. How long ago did the person die, to the nearest year.<BR>\n" );
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document.write( "The half-life decay equation that I am familiar with:<BR>\n" );
document.write( "  A = Ao*2^(-t/h)<BR>\n" );
document.write( "Where: <BR>\n" );
document.write( "A = resulting amt<BR>\n" );
document.write( "Ao = Original amt<BR>\n" );
document.write( "t = time in yrs (5715 for carbon 14)<BR>\n" );
document.write( "h = half-life in yrs<BR>\n" );
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document.write( "In the this problem result can be given in a decimal of .246, so Ao = 1.00<BR>\n" );
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document.write( "So we have:<BR>\n" );
document.write( "2^(-t/5715) = .246\r<BR>\n" );
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document.write( "Using natural logs equiv of exponents we have:<BR>\n" );
document.write( "(-t/5715)*ln(2) = ln(.246)<BR>\n" );
document.write( "-t/5715(.693147) = -1.40242<BR>\n" );
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document.write( "Get rid of the denominator and the negatives, mult equation by -5715<BR>\n" );
document.write( ".693147t = -1.40242*-5715<BR>\n" );
document.write( ".693147t = + 8014.85<BR>\n" );
document.write( "t = 8014.85/.693147<BR>\n" );
document.write( "t = 11,563 years<BR>\n" );
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document.write( "you can check this on a good calc; Enter 2^(-11563/5715) = .2459996<BR>\n" );
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document.write( "make sense to you? any questions?\r<BR>\n" );
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document.write( " </SPAN>\n" );
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