I was helping a friend do a math problem, and I was pretty stumped as to why we couldn't get to the right answer
Casey consumes an energy drink that contains caffeine. After consuming the energy drink, the amount of caffeine in Casey's body decreases exponentially. The 10-hour decay factor for the number of mg of caffeine in Casey's body is 0.2785.
Question c says:
If there were 173 mg of caffeine in Casey's body 1.56 hours after consuming the energy drink, how many mg of caffeine is in casey's body after 2.56 hours after consuming the energy drink?
So, the first thing I did was calculate the 1.56-hour decay factor, which should be $${0.2785}^{(1.56/10)}$$ Which is $0.819$. So this is where I was unsure which formula to use, as there are two:
$$N(t) = N_0(e^{-rt})$$ and $$f(t) = f_0 * (1-r)^t$$
So, I used both.
$$173 = N_0(e^{-0.819*1.56})$$ $$N_0 = 620.951$$ $$f(2.56)=620.951*(e^{-0.2785^{2.56/10}*2.56})$$ $$=98.0764$$
and
$$173 = N_0(1-0.2785^{(1.56/10)})^1.56$$ $$N_0 = 2493.65$$ $$f(2.56) = 2493.65(1-0.2785^{(2.56/10)})^{2.56}$$ $$=95.0563$$
Both of these answers are wrong. Am I doing something stupid here or what...?
You have
$$N(t) = N_0\left(e^{-rt}\right) \tag{1}\label{eq1A}$$
as a general Exponential decay equation where $N_0$ is the amount at $t = 0$, $N(t)$ is the amount remaining at time $t$ and $r$ is the exponential decay constant. You're specifically given that after $10$ hours, the decay factor is $0.2785$, i.e.,
$$\frac{N(10)}{N(0)} = \frac{N_0\left(e^{-10r}\right)}{N_0\left(e^{0}\right)} = e^{-10r} = 0.2785 \tag{2}\label{eq2A}$$
Taking the last $2$ parts of \eqref{eq2A} to the power of $0.1t$ gives
$$e^{-rt} = 0.2785^{0.1t} \tag{3}\label{eq3A}$$
This means that
$$N(t) = N_0\left(e^{-rt}\right) = N_0\left(0.2785^{0.1t}\right) \tag{4}\label{eq4A}$$
Also,
$$\frac{N(2.56)}{N(1.56)} = \frac{N_0\left(0.2785^{0.1(2.56)}\right)}{N_0\left(0.2785^{0.1(1.56)}\right)} = 0.2785^{0.1(2.56 - 1.56)} = 0.2785^{0.1} \tag{5}\label{eq5A}$$
Can you finish the rest?