Understanding the problem requires us to work with the continuous exponential decay model for a substance. Here is the step-by-step solution:
### Part (a)
Given:
- Initial amount of the substance ([tex]\(y_{initial}\)[/tex]) = 4300 kg
- Amount after 16 hours ([tex]\(y_{16}\)[/tex]) = 3053 kg
- Time ([tex]\(t_{16}\)[/tex]) = 16 hours
We need to find the exact formula relating [tex]\(y\)[/tex] to [tex]\(t\)[/tex]. The general formula for continuous exponential decay is:
[tex]\[ y = y_{initial} \cdot e^{kt} \][/tex]
To find the decay constant [tex]\(k\)[/tex], we use the formula at [tex]\(t = 16\)[/tex]:
[tex]\[ y_{16} = y_{initial} \cdot e^{k \cdot t_{16}} \][/tex]
This can be rearranged to:
[tex]\[ 3053 = 4300 \cdot e^{16k} \][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[ e^{16k} = \frac{3053}{4300} \][/tex]
Taking the natural logarithm on both sides:
[tex]\[ 16k = \ln{\left(\frac{3053}{4300}\right)} \][/tex]
[tex]\[ k = \frac{\ln{\left(\frac{3053}{4300}\right)}}{16} \][/tex]
Let’s denote this exact expression for [tex]\(k\)[/tex]:
[tex]\[ k = \frac{\ln{\left(\frac{3053}{4300}\right)}}{16} \][/tex]
Now we can write the formula for [tex]\(y\)[/tex] in terms of [tex]\(t\)[/tex]:
[tex]\[ y = 4300 \cdot e^{\left(\frac{\ln{\left(\frac{3053}{4300}\right)}}{16}\right) t} \][/tex]
### Part (b)
To find the amount of substance left after 18 hours, we use the formula derived:
[tex]\[ y = 4300 \cdot e^{\left(\frac{\ln{\left(\frac{3053}{4300}\right)}}{16}\right) \cdot 18} \][/tex]
First calculate the exponent:
[tex]\[ \text{Exponent} = \left(\frac{\ln{\left(\frac{3053}{4300}\right)}}{16}\right) \cdot 18 \][/tex]
Then, the amount of substance left after 18 hours:
[tex]\[ y_{18} = 4300 \cdot e^{\text{Exponent}} \][/tex]
By substituting the exact value obtained for the exponent:
[tex]\[ y_{18} \approx 2925.0554130428122 \][/tex]
Rounded to the nearest whole number:
[tex]\[ y_{18} \approx 2925 \][/tex]
So, the amount of substance left 18 hours after the beginning of the experiment is about 2925 kg.
Therefore, the complete answers are:
(a) The formula relating [tex]\(y\)[/tex] to [tex]\(t\)[/tex] is:
[tex]\[ y = 4300 \cdot e^{\left(\frac{\ln{\left(\frac{3053}{4300}\right)}}{16}\right) t} \][/tex]
(b) The amount of the substance left after 18 hours is approximately:
[tex]\[ 2925 \, \text{kg} \][/tex]
