SOLUTION: The amount C of cobalt-60 (in grams) in a storage facility at time t is given by C(t) = 25e^-.14t where time is measured in years. How long will it take for the original amo

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: The amount C of cobalt-60 (in grams) in a storage facility at time t is given by C(t) = 25e^-.14t where time is measured in years. How long will it take for the original amo      Log On


   

Question 948784: The amount C of cobalt-60 (in grams) in a storage facility at time t is given by C(t) = 25e^-.14t
where time is measured in years.
How long will it take for the original amount of 25 grams of the cobalt-60 to decay to half this amount. (This time is known as the half-life of cobalt-60.)
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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The amount C of cobalt-60 (in grams) in a storage facility at time t is given by C(t) = 25e^-.14t
where time is measured in years.
How long will it take for the original amount of 25 grams of the cobalt-60 to decay to half this amount. (This time is known as the half-life of cobalt-60.)
25%2Ae%5E%28-.14t%29+=+12.5
e%5E%28-.14t%29+=+12.5%2F25
e%5E%28-.14t%29+=+.5
use the nat logs
ln%28e%5E%28-.14t%29%29+=+%28.5%29
log equiv of exponents
-.14t%2Aln%28e%29+=+ln%28.5%29
Find the ln of .5 (the ln of e is 1)
-.14t = -.693
t = %28-.693%29%2F%28-.14%29
t = 4.95 yrs
:
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Check this on your calc: enter 25*e^(-.14*4.95)