document.write( "Question 63050: amount of a radioactive tracer remaining after t days is given by A=A0e^-0.058t, where A0 is the starting amount at the beginning of the time period. how many days will it take for one half of the original amount of decay? \n" ); document.write( "

Algebra.Com's Answer #43897 by funmath(2933) $\"\"$ $\"About$

You can put this solution on YOUR website!
amount of a radioactive tracer remaining after t days is given by A=Aoe^-0.058t, where Ao is the starting amount at the beginning of the time period. how many days will it take for one half of the original amount of decay?
\n" ); document.write( "A=(1/2)Ao
\n" ); document.write( " $\"%281%2F2%29Ao=Aoe%5E%28-0.058t%29\"$
\n" ); document.write( " $\"%281%2F2%29Ao%2FAo=Aoe%5E%28-0.058t%29%2FAo\"$
\n" ); document.write( " $\"1%2F2=e%5E%28-0.058t%29\"$
\n" ); document.write( " $\"ln%281%2F2%29=ln%28e%5E%28-0.058t%29%29\"$
\n" ); document.write( " $\"ln%281%2F2%29=-0.058tln%28e%29\"$ ln(e)=1
\n" ); document.write( " $\"ln%281%2F2%29=-0.058t%281%29\"$
\n" ); document.write( " $\"ln%281%2F2%29%2F-0.058=-0.058t%2F-0.058\"$
\n" ); document.write( " $\"-ln%281%2F2%29%2F0.058=t\"$
\n" ); document.write( " $\"t=11.95081346\"$
\n" ); document.write( "About 12 days.
\n" ); document.write( "Happy Calculating!!! \n" ); document.write( "

\n" );