document.write( "Question 173430This question is from textbook introduction to college mathematics
\n" ); document.write( ": The decay rate of krypton-85 is 6.3% per day. What is the half-life?
\n" ); document.write( "Using the exponential decay function P(t) =Po^e^-kt \n" ); document.write( "
Algebra.Com's Answer #128347 by ankor@dixie-net.com(22740)\"\" \"About 
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The decay rate of krypton-85 is 6.3% per day. What is the half-life?
\n" ); document.write( "Using the exponential decay function P(t) =Po^e^-kt
\n" ); document.write( ":
\n" ); document.write( "The exponential decay formula that I am familiar with:
\n" ); document.write( "A = Ao[2^(-t/h)]
\n" ); document.write( "where
\n" ); document.write( "Ao is the initial amt
\n" ); document.write( "A = final amt
\n" ); document.write( "t = time
\n" ); document.write( "h = half-life of the substance
\n" ); document.write( ":
\n" ); document.write( "Assume Ao = 100, Assume t = 1 day, find h (half life in days)
\n" ); document.write( "After 1 day, A = 100 - 6.3 = 93.7
\n" ); document.write( ":
\n" ); document.write( "100 * 2^(-1/h) = 93.7
\n" ); document.write( ":
\n" ); document.write( "2^(-1/h) = \"93.7%2F100\"
\n" ); document.write( ":
\n" ); document.write( "2^(-1/h) = .937
\n" ); document.write( ":
\n" ); document.write( "ln[2^(-1/h)) = ln(.937)
\n" ); document.write( ":
\n" ); document.write( "\"-1%2Fh\"*.693 = -.065
\n" ); document.write( ":
\n" ); document.write( "\"-.693%2Fh\" = -.065
\n" ); document.write( ":
\n" ); document.write( "h = \"%28-.693%29%2F%28-.065%29\"
\n" ); document.write( "h = 10.66 days is the half life of krypton-85 according to this information
\n" ); document.write( ":
\n" ); document.write( "In actual fact I think they mean 6.3% per year, hence, 10.66 yrs half life \n" ); document.write( "
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