Let's solve the given problem step-by-step.
### Part (a)
We are asked to write a formula relating the amount [tex]\( y \)[/tex] of the substance to the time [tex]\( t \)[/tex].
The general formula for continuous exponential decay is:
[tex]\[ y = y_0 \cdot e^{kt} \][/tex]
Here:
- [tex]\( y_0 \)[/tex] is the initial amount of the substance.
- [tex]\( k \)[/tex] is the decay constant.
- [tex]\( t \)[/tex] is the time in hours.
Given:
- Initial amount, [tex]\( y_0 = 4300 \)[/tex] kg
- Amount after 16 hours, [tex]\( y(16) = 3053 \)[/tex] kg
We need to determine the decay constant [tex]\( k \)[/tex]. Using the information for [tex]\( t = 16 \)[/tex]:
[tex]\[ 3053 = 4300 \cdot e^{16k} \][/tex]
To find [tex]\( k \)[/tex], solve the equation:
[tex]\[ e^{16k} = \frac{3053}{4300} \][/tex]
[tex]\[ e^{16k} = \frac{3053}{4300} \][/tex]
Taking the natural logarithm (ln) on both sides:
[tex]\[ 16k = \ln\left(\frac{3053}{4300}\right) \][/tex]
[tex]\[ k = \frac{1}{16} \ln\left(\frac{3053}{4300}\right) \][/tex]
Using the given result for precision:
[tex]\[ k = \log\left(10^{7/8} \cdot 71^{1/16} / 10\right) - \frac{7\pi i}{8} \][/tex]
Thus, the formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex] is:
[tex]\[ y = 4300 \cdot e^{ t \left( \log \left(10^{7/8} \cdot 71^{1/16} / 10 \right) - \frac{7\pi i}{8} \right) } \][/tex]
So, filling in the parts:
[tex]\[ y = 4300 \cdot e^{t (\log(10^{7/8} \cdot 71^{1/16} / 10) - 7\pi i / 8)} \][/tex]
### Part (b)
We need to find the amount remaining 18 hours after the beginning of the experiment.
Using the exponential decay formula with [tex]\( t = 18 \)[/tex]:
[tex]\[ y = 4300 \cdot e^{18k} \][/tex]
Where [tex]\( k \)[/tex] is the same decay constant we used earlier:
[tex]\[ k = \log(10^{7/8} \cdot 71^{1/16} / 10) - \frac{7\pi i}{8} \][/tex]
Substituting [tex]\( t = 18 \)[/tex] and the value of [tex]\( k \)[/tex]:
[tex]\[ y(18) = 4300 \cdot e^{18 (\log(10^{7/8} \cdot 71^{1/16} / 10) - 7\pi i / 8) } \][/tex]
From the result provided:
[tex]\[ y(18) = 3053 \cdot 10^{3/4} \cdot 71^{1/8} \cdot e^{i\pi/4} / 10 \][/tex]
[tex]\[ y(18) \approx 2068 + 2068i \][/tex]
After rounding to the nearest whole number, the amount of the substance remaining after 18 hours is approximately:
[tex]\[ 2068 \text{ kg (taking the real part)} \][/tex]
Thus, the amount of substance left 18 hours after the beginning of the experiment:
[tex]\[ 2068 \text{ kg} \][/tex]
