document.write( "Question 814289: A 1200 kg of a radioactive element kept decaying for 50 years until 300 kg of the element remained. What is the half-life of the element? I can not figure this answer out! Please help \n" ); document.write( "
Algebra.Com's Answer #490254 by ankor@dixie-net.com(22740)\"\" \"About 
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A 1200 kg of a radioactive element kept decaying for 50 years until 300 kg of the element remained.
\n" ); document.write( " What is the half-life of the element?
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\n" ); document.write( "The radioactive decay formula: A = Ao*2^(-t/h), WHERE:
\n" ); document.write( "A = Amt remaining after t time (300 mg)
\n" ); document.write( "Ao = initial amt (t=0) (1200 mg)
\n" ); document.write( "t = time of decay (50 yrs)
\n" ); document.write( "h = half-life of substance
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\n" ); document.write( "1200*2^(50/h) = 300
\n" ); document.write( "2^(-50/h) = \"300%2F1200\"
\n" ); document.write( "2^(-50/h) = .25
\n" ); document.write( "log[2^(-50/h)] = log(.25)
\n" ); document.write( "log equiv of exponents
\n" ); document.write( "\"-50%2Fh\"log(2) = log(.25)
\n" ); document.write( "\"-50%2Fh\" = \"log%28.25%29%2Flog%282%29\"
\n" ); document.write( "using your calc
\n" ); document.write( "\"-50%2Fh\" = -2
\n" ); document.write( "h = \"%28-50%29%2F%28-2%29\"
\n" ); document.write( "h = +25 yrs is the half life
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\n" ); document.write( ":
\n" ); document.write( "Check this on your calc, enter 1200*2^(-50/25) results: 300
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