document.write( "Question 449906: The half-life of carbon-14 is 5700 years.Find the age of a sample at which 18% of the radioactive nuclei originally present have decayed. \n" ); document.write( "
Algebra.Com's Answer #309591 by nerdybill(7384)\"\" \"About 
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The half-life of carbon-14 is 5700 years.Find the age of a sample at which 18% of the radioactive nuclei originally present have decayed.
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\n" ); document.write( "Exponential growth/decay formula:
\n" ); document.write( "N = No*e^(kt)
\n" ); document.write( "where
\n" ); document.write( "N is amount after time t
\n" ); document.write( "No is the initial amount
\n" ); document.write( "k is a constant
\n" ); document.write( "t is time
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\n" ); document.write( "Using:
\n" ); document.write( "The half-life of carbon-14 is 5700 years.
\n" ); document.write( "to find k:
\n" ); document.write( "Let x = initial amount
\n" ); document.write( "then
\n" ); document.write( ".5x = xe^(5700k)
\n" ); document.write( ".5 = e^(5700k)
\n" ); document.write( "ln(.5) = 5700k
\n" ); document.write( "ln(.5)/5700 = k
\n" ); document.write( "-.0001216 = k
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\n" ); document.write( "Now we can answer:
\n" ); document.write( "Find the age of a sample at which 18% of the radioactive nuclei originally present have decayed.
\n" ); document.write( "x-.18x = x*e^(-.0001216t)
\n" ); document.write( "x(1-.18) = x*e^(-.0001216t)
\n" ); document.write( "x(.82) = x*e^(-.0001216t)
\n" ); document.write( ".82 = e^(-.0001216t)
\n" ); document.write( "ln(.82) = -.0001216t
\n" ); document.write( "ln(.82)/(-.0001216) = t
\n" ); document.write( "1632 years = t\r
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