 # Joseph consumes an energy drink that contains caffeine. After consuming the energy drink, the amount of caffeine in Joseph's body decreases exponentia tricotasu 2020-11-08 Answered
Joseph consumes an energy drink that contains caffeine. After consuming the energy drink, the amount of caffeine in Josephs
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Given,
The 10-hour decay factor for the number of mg of caffeine in Joseph's body is 0.2601.
Exponential decay is the decrease in quantity N according to:
$N\left(t\right)={N}_{0}{e}^{-kt}$
where
${N}_{0}=$ initial value of quantity N
N(t) = quantity N at time t
k = decay constant associated to physical properties of N
${e}^{-kt}=$ decay factor
Substituting the values from the problem:
t = 10 hours
${e}^{-kt}=0.2601$
Then, solving for k:
Taking log on both side
$\mathrm{ln}\left({e}^{-kx10}\right)=\mathrm{ln}\left(0.2601\right)$
$-10k=\mathrm{ln}\left(0.2601\right)$
$k=\frac{-\mathrm{ln}\left(0.2601\right)}{10}$
k =0.13467
Therefore, for 10-hours decay factor = 0.13467
For t = 5 hours, the decay factor is given by:
${e}^{-kt}={e}^{-0.13467×5}$
$={e}^{-0.67335}$
= 0.5099
Therefore, for 5-hours decay factor =0.50999
For t = 1 hours, the decay factor is given by:
${e}^{-kt}={e}^{-0.13467×1}$
$={e}^{-0.13467}$
=0.87400
Therefore, for 1-hour decay factor =0.87400
If there are 166 mg in Joseph's body 1.38 hours after consuming the energy drink, then you could take this value as the initial value of quantity N0 Then, the quantity of caffeine in Joseph's body 2.38 hours later is just the quantity N(t) one hour later from the initial value (166 mg), then:
$N\left(t\right)={N}_{0}{e}^{-kt}$
N(1) $=166×0.87400$ (since for 1-hour decay factor =0.87400)
N(1)= 145.084 mg