if a radioactive element has a 5.23 days half-life and a beginning mass of 500g and has 3.90625g left. how old is the material and how many half-lives did it undergo?
OK, DC...we have this
3.90625 = 500 (1/2) ^(t/5.23) where t is in days
Divide both sides by 500
3.90625 / 500 = (1/2)^(t/5.23) take the log of both sides
log ( 3.90625 / 500) = log (1/2)^(t/5.23) and we can write
log (3.90625 / 500) = (t / 5.23) * log (1/2)
Multiply both sides by 5.23 / log (1/2)
log (3.90625 / 500) * (5.23 / log (1/2) ) = t ≈ 36.61 days
To find the number of half-lives we have
36.61 / 5.23 = 7 half-lives
Another take (but similar)
500 e^-kt = 250 to solve for k we will let t= number of half lives= 1 (as in ONE half life)
k = ln (250/500) / -1 k=.693147
Now for the sample
500 e^-.693147 t = 3.90625
t = #half lifes = ln (3.90625/5000) / - .693147 t = 7.00 half lifes (or is it half LIVES ?)
It’s “half-lives” with a hyphen. CPhil used it, but most mathematicians don’t use this because they think it means to subtract.
Part of the fun of figuring out the math problem is figuring out the written mush the answerer gives along with the solution.