document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=277451>Question 277451</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Radioactive carbon-14 decays according to the function Q(t)=Qoe^-0.000121t where t is time in years, Q(t) is the quantity remaining at time t, and Qo is the amount of present at time t=0. Estimate the age of a skull if 23% of the original quantity of carbon-14 remains.\r<BR>\n" );
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document.write( "Thank you! TT^TT </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #201985 by </B><A HREF=/tutors/aboutme.mpl?userid=nerdybill><B>nerdybill(7384)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=nerdybill><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=nerdybill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=201985\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Radioactive carbon-14 decays according to the function Q(t)=Qoe^-0.000121t where t is time in years, Q(t) is the quantity remaining at time t, and Qo is the amount of present at time t=0. Estimate the age of a skull if 23% of the original quantity of carbon-14 remains. <BR>\n" );
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document.write( "Q(t)=Qoe^-0.000121t<BR>\n" );
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document.write( "Plugging in:<BR>\n" );
document.write( ".23(Qo) = (Qo)e^-0.000121t<BR>\n" );
document.write( "Now, we solve for t<BR>\n" );
document.write( "first, divide both sides by Qo:<BR>\n" );
document.write( ".23 = e^-0.000121t<BR>\n" );
document.write( "take the natural log of both sides:<BR>\n" );
document.write( "ln(.23) = -0.000121t<BR>\n" );
document.write( "ln(.23)/(-0.000121) = t<BR>\n" );
document.write( "12,146 years = t<BR>\n" );
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