How long, in years, will it take 80 kg of plutonium (half life 24 000 years) to decay to approximately 1.8 kg?
solve \[1.8=80\times (\frac{1}{2})^{\frac{t}{24,000}}\] for \(t\)
first divide both sides by 80 to get \[.0225=(\frac{1}{2})^{\frac{t}{24,000}}\] then change of base gives \[\frac{t}{24,000}=\frac{\ln(.0225)}{\ln(.5)}\] and multiply by 24,000 and use a calculator
okay and can u see if i have the right answer for this?
i got 228 years
no that cannot be right
if the half life is 24,000 years then that is the time it will take to go from 80 to 40
oh sorry this answer is for a different questionj and i wanted u to check if i got the answer right
looks like about 131317 http://www.wolframalpha.com/input/?i=24000*ln%28.0225%29%2Fln%28.5%29
260 grams of a radioactive substance decays to 160 grams after 160 years. To the nearest year, what is the half-life of the substance?
is the answer for this 228
\[160=260(\frac{1}{2})^t\] \[\frac{8}{13}=(\frac{1}{2})^t\] \[t=\frac{\ln(\frac{8}{13})}{\ln(.5)}\]
sorry if im asking too many questions. can u help me out with a few more if you dont mind?
damn i made a mistake. have to multiply my answer by 160
i get 112
so what ever you get for t you multiply it by 160?
r u sure?
equation at the beginning should have been \[160=260\times (\frac{1}{2})^{\frac{t}{160}}\]
can u help me with this one? A stock that Jane bought at $75 a share is now worth $629.86 a share. If it was averaged a 12% increase in value a year, then how many years has Jane owned the stock? Round off your answer to one decimal place.
solution method is identical, but you should end up with \[\frac{t}{160}=\frac{\ln(\frac{8}{13})}{\ln(.5)}\] \[t=160\times \frac{\ln(\frac{8}{13})}{\ln(.5)}\]\]
\[629.86=75(1.12)^t\] solve for t
so would it be ln629.86/ ln 75
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