document.write( "Question 194226: a piece of bone from an organism is found to contain 10% of the carbon -14 is 5730 years then how long ago was the organism was alive?\r
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document.write( "please help me solve!!!!!\r
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document.write( "i really dont know how tio solve pleaseeeee help..... \n" );
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Algebra.Com's Answer #145771 by scott8148(6628) You can put this solution on YOUR website! the half-life of carbon-14 is 5730yr \n" ); document.write( "this means that in 5730yr, 50% of the original amount will remain \n" ); document.write( "in another 5730yr, 25% will remain \n" ); document.write( "and so on\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the number of half-lives can be found from ___ p = (1/2)^n \n" ); document.write( "where p is the fraction remaining and n is the number of half-lives\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "taking the log ___ log(p) = n * log(1/2) \n" ); document.write( "dividing by log(1/2) ___ [log(p)] / [log(1/2)] = n\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "[log(.10)] / [log(1/2)] = n \n" ); document.write( "using a calculator ___ n = 3.32 (approx)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "so the time is ___ 3.32 * 5730 = 19024yr \n" ); document.write( " |