document.write( "Question 1201597: Radioactive Carbon-14,which is found in all living things, has a half-life of 5700 years and can be used to date fossils. Suppose a fossil contains 15% of the amount of Carbon-14 that the organism contained when it was alive. Calculate the age of the fossil. \n" ); document.write( "
Algebra.Com's Answer #836052 by math_tutor2020(3817)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Answer: 15600 years old (approximate)\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Work Shown:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "a = starting amount = doesn't matter since it cancels out later \n" ); document.write( "0.15a = 15% of the starting amount\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "x = number of years \n" ); document.write( "y = amount of material remaining \n" ); document.write( "H = 5700 = half-life\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Half-life formula \n" ); document.write( "y = a*(0.5)^(x/H) \n" ); document.write( "0.15a = a*(0.5)^(x/5700) \n" ); document.write( "0.15 = (0.5)^(x/5700) \n" ); document.write( "Log(0.15) = Log( (0.5)^(x/5700) ) \n" ); document.write( "Log(0.15) = (x/5700)*Log(0.5) \n" ); document.write( "x/5700 = Log(0.15)/Log(0.5) \n" ); document.write( "x = 5700*Log(0.15)/Log(0.5) \n" ); document.write( "x = 15600.7038867473 \n" ); document.write( "x = 15600\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The substance is roughly 15600 years old. \n" ); document.write( "Round this however your teacher instructs. \n" ); document.write( "I decided to round to the nearest hundred. \n" ); document.write( " \n" ); document.write( " |