document.write( "Question 314580: 3. An isotope of sodium, Na^24, has a half-life of 15 hours. How many hours will it take to decay to 2^(-4) grams, if the initial mass is 4 grams? \n" ); document.write( "
Algebra.Com's Answer #224960 by nerdybill(7384)\"\" \"About 
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3. An isotope of sodium, Na^24, has a half-life of 15 hours. How many hours will it take to decay to 2^(-4) grams, if the initial mass is 4 grams?
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\n" ); document.write( "Exponential growth/decay formula:
\n" ); document.write( "A = Ne^(rt)
\n" ); document.write( "where
\n" ); document.write( "A is amount after time t
\n" ); document.write( "N is the initial amount
\n" ); document.write( "r is the rate of growth/decay
\n" ); document.write( "t is time
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\n" ); document.write( "Since we know half-life is 15 hours
\n" ); document.write( "Let N = initial amount
\n" ); document.write( "then
\n" ); document.write( "A must by N/2 giving you
\n" ); document.write( "N/2 = Ne^(15r)
\n" ); document.write( "Solving for r:
\n" ); document.write( "1/2 = e^(15r)
\n" ); document.write( "ln(1/2) = 15r
\n" ); document.write( "ln(1/2)/15 = r
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\n" ); document.write( "Our formula is now:
\n" ); document.write( "A = Ne^(tln(1/2)/15)
\n" ); document.write( "We can now use the above to answer:
\n" ); document.write( "How many hours will it take to decay to 2^(-4) grams, if the initial mass is 4 grams?
\n" ); document.write( "2^(-4) = 4e^(tln(1/2)/15)
\n" ); document.write( "2^(-4)/4 = e^(tln(1/2)/15)
\n" ); document.write( "ln(2^(-4)/4) = tln(1/2)/15
\n" ); document.write( "ln(2^(-4)/4)/(ln(1/2)/15) = t
\n" ); document.write( "ln(.015625)/(ln(1/2)/15) = t
\n" ); document.write( "90 hours = t\r
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