document.write( "Question 231367: Th half-life of Carbon 14 is 5730 years. If 8 grams remain after 1000 years, what was the initial quantity? \n" ); document.write( "
Algebra.Com's Answer #171263 by nerdybill(7384)\"\" \"About 
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\n" ); document.write( "Th half-life of Carbon 14 is 5730 years. If 8 grams remain after 1000 years, what was the initial quantity?
\n" ); document.write( ".
\n" ); document.write( "Exponential decay is described by:
\n" ); document.write( "N(t) = N(0) e^(kt)
\n" ); document.write( "where
\n" ); document.write( "N(t) amount at time t
\n" ); document.write( "N(0) initial amount
\n" ); document.write( "k is the growth rate
\n" ); document.write( "t is time
\n" ); document.write( ".
\n" ); document.write( "First we need to figure out what k is...
\n" ); document.write( "x/2 = x * e^(5730k)
\n" ); document.write( "1/2 = e^(5730k)
\n" ); document.write( "ln(1/2) = 5730k
\n" ); document.write( "ln(1/2)/5730 = k
\n" ); document.write( ".
\n" ); document.write( "Now, we can answer:
\n" ); document.write( "If 8 grams remain after 1000 years, what was the initial quantity?
\n" ); document.write( "N(t) = N(0) e^(kt)
\n" ); document.write( "Let x = initial quantity
\n" ); document.write( "8 = x * e^(1000 * ln(1/2)/5730)
\n" ); document.write( "8/e^(1000 * ln(1/2)/5730) = x
\n" ); document.write( "9.03 grams = x\r
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