document.write( "Question 812496: the half life of radium is 1690 years. if 60 grams are present now, how much will be present in 670 years? \n" ); document.write( "
Algebra.Com's Answer #489175 by josgarithmetic(39617)\"\" \"About 
You can put this solution on YOUR website!
Basic General Model, \"A=Ao%2Ae%5E%28kt%29\"
\n" ); document.write( "\"ln%28A%29=ln%28Ao%29%2Bkt%2Aln%28e%29\"
\n" ); document.write( "\"ln%28A%29=ln%28Ao%29%2Bkt\"
\n" ); document.write( "\"k=%28ln%28A%29-ln%28Ao%29%29%2Ft\"
\n" ); document.write( "\"k=%28ln%281%2F2%29-ln%281%29%29%2F1690\"
\n" ); document.write( "\"highlight%28k=-0.000952%29\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"A=60%2Ae%5E%28-0.000952%2At%29\"
\n" ); document.write( "\"A=60%2Ae%5E%28-0.000952%2A670%29\"
\n" ); document.write( "\"highlight%28A=32%29\", close to 31.7 grams.... This is a mistake...\r
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\n" ); document.write( "\n" ); document.write( "CHECK:
\n" ); document.write( "\"1%2F2=1%2Ae%5E%28kt%29\"
\n" ); document.write( "\"ln%280.5%29=kt%2Aln%28e%29\"
\n" ); document.write( "\"kt=ln%280.5%29\"
\n" ); document.write( "\"k=%281%2F1690%29ln%280.5%29=-0.0004101\" as the constant k. Believing this to be a more correct value, you should find \"A=60%2Ae%5E%28-0.0004101%2A670%29\"
\n" ); document.write( "\"highlight%28A=45%29\", close to 45.4 grams. The work shown above \"CHECK:\" contains a mistake.
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