document.write( "Question 1048723: the half-life of a certain radioactive substance is 8 hours. there are 5 grams present initially. which of the following gives best approximation when there will be 1 gram remaining. \n" ); document.write( "
Algebra.Com's Answer #664333 by josgarithmetic(39617)\"\" \"About 
You can put this solution on YOUR website!
\"y=p%2Ae%5E%28-kt%29\"--------exponential decay model\r
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\n" ); document.write( "\n" ); document.write( "Need find k. The variable to find, AFTER k is found, will be t.\r
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\n" ); document.write( "\n" ); document.write( "HALF LIFE
\n" ); document.write( "\"p%2F2=p%2Ae%5E%28-k%2A8%29\"
\n" ); document.write( "\"%281%2F2%29=e%5E%28-8k%29\"
\n" ); document.write( "\"ln%281%2F2%29=ln%28e%5E%28-8k%29%29\"
\n" ); document.write( "\"-8k%2Aln%28e%29=ln%281%2F2%29\"
\n" ); document.write( "\"-8k=-ln%282%29\"
\n" ); document.write( "\"highlight_green%28k=ln%282%29%2F8%29\"
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\n" ); document.write( "You would like to put a decimal number to that.
\n" ); document.write( "\"k=0.08664\", but remember to include the negative sign in the model!!\r
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\n" ); document.write( "\n" ); document.write( "WHEN THE 5 GRAMS BECOME 1 GRAM? FIND t.
\n" ); document.write( "\"ln%28y%29=ln%28p%2Ae%5E%28-kt%29%29\"
\n" ); document.write( "\"ln%28p%29-kt=ln%28y%29\"
\n" ); document.write( "\"-kt=ln%28y%29-ln%28p%29\"
\n" ); document.write( "\"-kt=ln%28y%2Fp%29\"
\n" ); document.write( "\"t=%28-1%2Fk%29ln%28y%2Fp%29\"
\n" ); document.write( "\"highlight%28t=%281%2Fk%29ln%28p%2Fy%29%29\", if you want to keep the positive value of k found earlier.\r
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\n" ); document.write( "\n" ); document.write( "YOU FINISH IT.
\n" ); document.write( "\"system%28p=5%2Cy=1%2Ck=0.08664%2Ct=YouNeedToEvaluateHours%29\"
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