The amount of caffeine in the body after

To solve the problem, we can use the exponential decay model for the amount of caffeine in the body after consumption. Let's denote the initial amount of caffeine as $C_0$ and the amount of caffeine remaining after time $t$ as $C(t)$.
According to the problem, 50% of the caffeine is left in the body after approximately 6 hours. This means that after 6 hours, the remaining caffeine is half of the initial amount. Mathematically, we can express this as:
$C(6)=\frac{1}{2}{C}_0$
Now, we need to find the amount of caffeine remaining in the body after 9 hours, denoted as $C(9)$.
To solve for $C(9)$, we can use the exponential decay model. The general form of an exponential decay function is:
$C(t)=C_0·e^{-kt}$
where $k$ is the decay constant.
We can substitute the given information into the equation:
$\frac{1}{2}{C}_0=C_0·e^{-k·6}$
Canceling out $C_0$ on both sides:
$\frac{1}{2}=e^{-6k}$
To solve for $k$, we can take the natural logarithm (ln) of both sides:
$\ln \left(\frac{1}{2}\right)=\ln \left(e^{-6k}\right)$
Using the property $\ln (e^x)=x$, the equation simplifies to:
$\ln \left(\frac{1}{2}\right)=-6k$
Solving for $k$:
$k=-\frac{\ln \left(\frac{1}{2}\right)}{6}$
Now that we have the value of $k$, we can find the amount of caffeine remaining after 9 hours:
$C(9)=C_0·e^{-k·9}$
Substituting the value of $k$:
$C(9)=C_0·e^{-(-\frac{\ln \left(\frac{1}{2}\right)}{6})·9}$
Simplifying further:
$C(9)=C_0·e^{\frac{\ln \left(\frac{1}{2}\right)}{6}·9}$
Using the property $e^{\ln (x)}=x$, we have:
$C(9)=C_0·{\left(\frac{1}{2}\right)}^{\frac{9}{6}}$
Simplifying the exponent:
$C(9)=C_0·{\left(\frac{1}{2}\right)}^{\frac{3}{2}}$
Taking the square root:
$C(9)=C_0·{\left(\frac{1}{\sqrt{2}}\right)}^3$
Simplifying further:
$C(9)=C_0·\frac{1}{{\sqrt{2}}^3}$
$C(9)=C_0·\frac{1}{8\sqrt{2}}$
Therefore, the amount of caffeine left in your system after 9 hours is $\frac{1}{8\sqrt{2}}$ times the initial amount $C_0$.