document.write( "Question 966341: An initial amount of a radioactive substance y0 is given, along with information about the amount remaining after a given time t in appropriate units. For an equation of the form y = y0ekt that models the situation, give the exact value of k in terms of natural logarithms: y0 = 20 mg; the half-life is 200 days. \n" ); document.write( "
Algebra.Com's Answer #590661 by ankor@dixie-net.com(22740)\"\" \"About 
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An initial amount of a radioactive substance y0 is given, along with information about the amount remaining after a given time t in appropriate units.
\n" ); document.write( " For an equation of the form y = y0ekt that models the situation, give the exact value of k in terms of natural logarithms: y0 = 20 mg; the half-life is 200 days.
\n" ); document.write( ":
\n" ); document.write( "The equation is:
\n" ); document.write( "Half remains after 200 days
\n" ); document.write( "\"Y+=+Yo%2Ae%5E%28kt%29\" where:
\n" ); document.write( "Y = amt after t time (10 mg)
\n" ); document.write( "Yo = initial amt (20 mg)
\n" ); document.write( "k = half-life of the substance
\n" ); document.write( "t = time of decay (200)
\n" ); document.write( ":
\n" ); document.write( "\"20%2Ae%5E%28200k%29+=+10\"
\n" ); document.write( "\"e%5E%28200k%29+=+10%2F20\"
\n" ); document.write( "\"e%5E%28200k%29+=+1%2F2\"
\n" ); document.write( "ln of e is 1, therefore
\n" ); document.write( "\"200k+=+ln%281%2F2%29\"
\n" ); document.write( "200k = -.693147
\n" ); document.write( "k = \"-.693147%2F200\"
\n" ); document.write( "k = -.0034657
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check this on your calc: 20*e^(-.0034657*200) = 10.000
\n" ); document.write( "
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