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A 29 gram sample of a substance thats used for drug research has a k-value of 0.1359. N=Noe^-kt No=initial mass( at time t=0) N=mass at time t k=a positive constant that depends on the substance itself and on the units used to measure time t=time, in days Find the substance's half-life,in days.Round your answer to the nearest tenth.
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\[n=n_0e^{-kt}\\\]
im still confused
\[n=n_0 e^{-0.1359t}\\n=0 \rightarrow n_0=n_0 e^0=29 gr\\n_0=29\\n=29* e^{-0.1359t}\]
for a half life solve n=n0/2 \[n=29* e^{-0.1359t}=\frac{n_0}{2}=\frac{29}{2}\\ e^{-0.1359t}=\frac{1}{2}\\ln\\ln( e^{-0.1359t})=\ln(\frac{1}{2})\\0.1359t=-0.6931\\t=5.100\]
\[t \approx 5\]
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