document.write( "Question 324058: A radioactive substance has a half-life of 6.8 hours. If 12.6 kg. remain from the original amount of 20 kg. How many hours have passed? \n" ); document.write( "
Algebra.Com's Answer #231913 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
A radioactive substance has a half-life of 6.8 hours.
\n" ); document.write( " If 12.6 kg. remain from the original amount of 20 kg.
\n" ); document.write( " How many hours have passed?
\n" ); document.write( ":
\n" ); document.write( "The half-life formula: A = Ao*2^(-t/h)
\n" ); document.write( "Where
\n" ); document.write( "A = Resulting amt after t hrs
\n" ); document.write( "Ao = initial amt
\n" ); document.write( "t = time (in hrs)
\n" ); document.write( "h = half-life of the substance
\n" ); document.write( ":
\n" ); document.write( "20*2^(-t/6.8) = 12.6
\n" ); document.write( ":
\n" ); document.write( "2^(-t/6.8) = \"12.6%2F20\"
\n" ); document.write( "2^(-t/6.8) = .63
\n" ); document.write( "use logs here
\n" ); document.write( "log(2^(-t/6.8)) = log(.63)
\n" ); document.write( "log equiv of exponents
\n" ); document.write( "\"-t%2F6.8\"*log(2) = log(.63)
\n" ); document.write( "Find the logs of 2 and .63
\n" ); document.write( "\"-t%2F6.8\"*.301 = -.201
\n" ); document.write( "\"-.301t%2F6.8\" = -.201
\n" ); document.write( ":
\n" ); document.write( "-.301t = -.201 * 6.8
\n" ); document.write( "-.301t = -1.3645
\n" ); document.write( "t = \"%28-1.3645%29%2F%28-.301%29\"
\n" ); document.write( "t = 4.5 hrs, for 12.6 kg to remain
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check solution on a calc; enter: 20*2^(4.5/6.8) results 12.6 \n" ); document.write( "
\n" );