document.write( "Question 1159088: A wooden artifact from an ancient tomb contains 35 percent of the carbon-14 that is present in living trees.\r
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Algebra.Com's Answer #782079 by ikleyn(52786)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "In terms of the half-life, the general formula for radioactive decay of the Carbon-14 is\r\n" );
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document.write( "    C(t) = \"C%280%29\".\"%281%2F2%29%5E%28t%2F5730%29\"\r\n" );
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document.write( "where C(t) is the current mass of the carbon-14; C(0) is the initial mass,\r\n" );
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document.write( "Since 335% of the Carbon-14 remained, you have this equation\r\n" );
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document.write( "    0.35*C(0) = \"C%280%29\".\"%281%2F2%29%5E%28t%2F5730%29\",  which reduces to   0.35 = \"%281%2F2%29%5E%28t%2F5730%29\",\r\n" );
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document.write( "To solve it, take logarithm base 10 from both sides. You get an equation \r\n" );
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document.write( "    \"log%2810%2C%280.35%29%29\" = \"log%2810%2C%281%2F2%29%5E%28t%2F5730%29%29%29\",  or  \"%28t%2F5730%29%2Alog%2810%2C%280.5%29%29\" = \"log%2810%2C%280.35%29%29\".\r\n" );
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document.write( "Therefore,\r\n" );
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document.write( "     t = \"5730%2A%28log%2810%2C%280.35%29%29%2Flog%2810%2C%280.5%29%29%29\" = 8768 years.\r\n" );
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document.write( "ANSWER.  The piece of wood is about 8768 years old.\r\n" );
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\n" ); document.write( "\n" ); document.write( "The post-solution note:\r
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document.write( "    Since the half-life parameter is given, it is NATURALLY to have the ENTIRE SOLUTION in half-life terms.\r\n" );
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document.write( "    The transition to other form (to \"ekt-form\"), as @josgaritmetic starts his solution, is not necessary.\r\n" );
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document.write( "    It only leads to unnecessary excessive job and creates the room for errors.\r\n" );
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