document.write( "Question 688702: Suppose that 5g of a radioactive substance decrease to 4g in 30 seconds . How long does it take for 3/5 of the substance to decay? \n" ); document.write( "
Algebra.Com's Answer #425808 by ankor@dixie-net.com(22740)\"\" \"About 
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Suppose that 5g of a radioactive substance decrease to 4g in 30 seconds .
\n" ); document.write( "find the half-life of the this substance
\n" ); document.write( "A = Ao*2^(-t/h), where
\n" ); document.write( "A = amt after t time
\n" ); document.write( "Ao = initial amt
\n" ); document.write( "t = time of decay
\n" ); document.write( "h = half-life of substance
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\n" ); document.write( "5*2^(-30/h) = 4
\n" ); document.write( "2^(-30/h) = 4/5
\n" ); document.write( "2^(-30/h) = .8
\n" ); document.write( "\"-30%2Fh\"ln(2) = ln(.8)
\n" ); document.write( "\"-30%2Fh\" = \"ln%28.8%29%2Fln%282%29\"
\n" ); document.write( "\"-30%2Fh\" = -.322
\n" ); document.write( "h = \"%28-30%29%2F%28-.322%29%29\"
\n" ); document.write( "t ~ 93 seconds is the half life of the substance
\n" ); document.write( ":
\n" ); document.write( " How long does it take for 3/5 of the substance to decay?
\n" ); document.write( "3/5 decay of 5g leaves 2g remaining
\n" ); document.write( "5*2^(-t/93) = 2
\n" ); document.write( "2^(-t/93) = 2/5
\n" ); document.write( "2^(-t/93) = .4
\n" ); document.write( "\"-t%2F93\" = \"ln%28.4%29%2Fln%282%29\"
\n" ); document.write( "\"-t%2F93\" = -1.322
\n" ); document.write( "t = -93 * -1.322
\n" ); document.write( "t ~ 123 seconds\r
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