document.write( "Question 1199250: The half-life of a radioactive substance is one hundred twenty-five days. How many days will it take for eighty-two percent of the substance to decay?\r
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Algebra.Com's Answer #833036 by Edwin McCravy(20059)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "The formula is\r\n" );
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document.write( "\"A=A%5Bo%5De%5E%28kt%29\", where Ao is the beginning amount, A is the final amount,\r\n" );
document.write( "t is time required to reach the final amount, and k is a constant.\r\n" );
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document.write( "If we start with quantity P, then when t = 25, the final amount A is half\r\n" );
document.write( "the starting amount P, so \"A=expr%281%2F2%29P\".\r\n" );
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document.write( "\"expr%281%2F2%29P=Pe%5E%28k%2A25%29\"\r\n" );
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document.write( "Cancel the P's\r\n" );
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document.write( "\"expr%281%2F2%29cross%28P%29=cross%28P%29e%5E%28kt%29\" \r\n" );
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document.write( "\"1%2F2=e%5E%28k%2A125%29\"\r\n" );
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document.write( "Take natural logs of both sides:\r\n" );
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document.write( "\"ln%281%2F2%29=ln%28e%5E%28k%2A125%29%29\"\r\n" );
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document.write( "\"-0.6931471806=k%2A125\"\r\n" );
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document.write( "\"-0.6931471806%2F125=k%2A125%2F125\"\r\n" );
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document.write( "\"-0.0055451774=k\"\r\n" );
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document.write( "Now we go back to \r\n" );
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document.write( "\"A=Pe%5E%28kt%29\"\r\n" );
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document.write( "and replace k by -0.0055451774 \r\n" );
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document.write( "\"A=Pe%5E%28-0.0055451774t%29\"\r\n" );
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document.write( "Now we must find how many days it will take for eighty-two percent \r\n" );
document.write( "of the substance to decay?\r\n" );
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document.write( "Remember, A is how much is left, not how much decaying has been\r\n" );
document.write( "done, so when 82% of the stuff decays, what's left is 100%-82% = 18%. \r\n" );
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document.write( "So we substitute 0.18P for A\r\n" );
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document.write( "\"0.18P=Pe%5E%28-0.0055451774t%29\"\r\n" );
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document.write( "Cancel P's\r\n" );
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document.write( "\"0.18cross%28P%29=cross%28P%29e%5E%28-0.0055451774t%29\"\r\n" );
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document.write( "\"0.18=e%5E%28-0.0055451774t%29\"\r\n" );
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document.write( "\"ln%280.18%29=ln%28e%5E%28-0.0055451774t%29%29\"\r\n" );
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document.write( "\"-1.714798428=-0.0055451774t\"\r\n" );
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document.write( "\"%28-1.714798428%29%2F%28-0.0055451774%29=t\"\r\n" );
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document.write( "\"309.241401=t\"\r\n" );
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document.write( "Not quite all 82% will have decayed on the 309th day,\r\n" );
document.write( "but it will all will be decayed on the 310th day.\r\n" );
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document.write( "Edwin
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