document.write( "Question 646647: The half-life of radioactive strontium-90 is approximately 29 years. In 1964, radioactive strontium-90 was released into the atmosphere during testing of nuclear weapons, and was absorbed into people's bones. How many years (since 1964) does it take until only 9 percent of the original amount absorbed remains? \n" ); document.write( "
Algebra.Com's Answer #406008 by ankor@dixie-net.com(22740)\"\" \"About 
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The half-life of radioactive strontium-90 is approximately 29 years.
\n" ); document.write( " In 1964, radioactive strontium-90 was released into the atmosphere during
\n" ); document.write( "testing of nuclear weapons, and was absorbed into people's bones.
\n" ); document.write( " How many years (since 1964) does it take until only 9 percent of the
\n" ); document.write( " original amount absorbed remains?
\n" ); document.write( ":
\n" ); document.write( "The radio-active decay formula: A = Ao*2^(-t/h), where
\n" ); document.write( "A = resulting amt after t time
\n" ); document.write( "Ao = initial amt (t=0)
\n" ); document.write( "t = time of decay
\n" ); document.write( "h = half-life of substance
\n" ); document.write( ":
\n" ); document.write( "Let initial amt = 100 then resulting amt = 9
\n" ); document.write( "100*2^(-t/29) = 9
\n" ); document.write( "divide both sides by 100
\n" ); document.write( "2^(-t/29) = .09
\n" ); document.write( "using nat logs, the log equiv of exponents
\n" ); document.write( "\"-t%2F29\"*ln(2) = ln(.09)
\n" ); document.write( "\"-t%2F29\" = \"ln.09%2Fln%282%29\"
\n" ); document.write( "\"-t%2F29\" = -3.47393
\n" ); document.write( "t = -29 * -3.47393
\n" ); document.write( "t = 100.744 yrs, plus 1964 = 2065 for bones to be reduced to 9%
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "That seemed like a long time, checked on a calc:
\n" ); document.write( "enter 100*2^(-100.744/29) results: 9.000 \n" ); document.write( "
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