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A certain substance decomposes according to a continuous exponential decay model. To begin an experiment, the initial amount of the substance is 4300 kg. After 16 hours, 3053 kg of the substance is left.

(a) Let [tex] t [/tex] be the time (in hours) since the beginning of the experiment, and let [tex] y [/tex] be the amount of the substance (in kg) at time [tex] t [/tex]. Write a formula relating [tex] y [/tex] to [tex] t [/tex]. Use exact expressions to fill in the missing parts of the formula. Do not use approximations.

[tex]\[ y = 4300 e^{kt} \][/tex]

(b) How much of the substance is left 18 hours after the beginning of the experiment? Do not round any intermediate computations, and round your answer to the nearest whole number.

[tex]\[ \boxed{ \text{kg} } \][/tex]

Sagot :

GinnyAnsweridn
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]
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