Math / Calculus

QuestionThe population of a culture of bacteria, $P(t)$ , where $t$ is time in days, is growing at a rate that is proportional to the population itself and the growth rate is 0.4 . The initial population is 20 . (a.) What is the population after 30 days? Do not round your answer.
Population after 30 days = $\square$ (b.) How much time does it take for the population to double? Round your answer to the nearest tenth.
Time to double the population $\approx$ $\square$ days

Studdy Solution

STEP 1

What is this asking? We've got some bacteria multiplying like crazy, and we need to figure out how many there are after 30 days and how long it takes for their population to double. Watch out! Don't mix up the population itself with the rate at which it's growing!

STEP 2

1. Set up the equation
2. Calculate the population after 30 days
3. Calculate the time to double

STEP 3

We know the population growth is proportional to the current population.
This screams exponential growth!
The general formula for exponential growth is $P(t) = P_0 \cdot e^{kt}$ , where $P(t)$ is the population at time $t$ , $P_0$ is the initial population, $k$ is the growth rate, and $t$ is the time.

STEP 4

In our case, the initial population $P_0$ is 20, and the growth rate $k$ is 0.4.
So, our equation becomes $P(t) = 20 \cdot e^{0.4t}$ .
Awesome!

STEP 5

We want to find the population after 30 days, so we need to plug in $t = \textbf{30}$ into our equation: $P(30) = 20 \cdot e^{0.4 \cdot \textbf{30}}$ .

STEP 6

Now, let's calculate: $P(30) = 20 \cdot e^{12}$ .
This gives us $P(30) \approx 20 \cdot 162754.7914 = 3255095.828$ .
So, after 30 days, there will be approximately 3,255,095.828 bacteria!

STEP 7

We want to find the time it takes for the population to double.
That means we're looking for the time $t$ when $P(t) = 2 \cdot P_0$ , which is $2 \cdot 20 = 40$ .

STEP 8

Let's set up the equation: $40 = 20 \cdot e^{0.4t}$ .
To solve for $t$ , we can first divide both sides by 20: $\frac{40}{20} = \frac{20 \cdot e^{0.4t}}{20}$ which simplifies to $2 = e^{0.4t}$ .

STEP 9

Now, we can take the natural logarithm (ln) of both sides: $\ln(2) = \ln(e^{0.4t})$ .
This simplifies to $\ln(2) = 0.4t$ .

STEP 10

Finally, we can divide both sides by 0.4 to isolate $t$ : $t = \frac{\ln(2)}{0.4} \approx \frac{0.6931}{0.4} \approx 1.7328$ .
Rounding to the nearest tenth gives us $t \approx \textbf{1.7}$ days.

STEP 11

(a.) After 30 days, the population is 3,255,095.828 bacteria. (b.) It takes approximately 1.7 days for the population to double.

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Studdy Solution

STEP 1

What is this asking? We've got some bacteria multiplying like crazy, and we need to figure out how many there are after 30 days and how long it takes for their population to double. Watch out! Don't mix up the population itself with the rate at which it's growing!

STEP 2

1. Set up the equation
2. Calculate the population after 30 days
3. Calculate the time to double

STEP 3

STEP 4

In our case, the initial population $P_0$ is 20, and the growth rate $k$ is 0.4.
So, our equation becomes $P(t) = 20 \cdot e^{0.4t}$ .
Awesome!

STEP 5

We want to find the population after 30 days, so we need to plug in $t = \textbf{30}$ into our equation: $P(30) = 20 \cdot e^{0.4 \cdot \textbf{30}}$ .

STEP 6

Now, let's calculate: $P(30) = 20 \cdot e^{12}$ .
This gives us $P(30) \approx 20 \cdot 162754.7914 = 3255095.828$ .
So, after 30 days, there will be approximately 3,255,095.828 bacteria!

STEP 7

We want to find the time it takes for the population to double.
That means we're looking for the time $t$ when $P(t) = 2 \cdot P_0$ , which is $2 \cdot 20 = 40$ .

STEP 8

Let's set up the equation: $40 = 20 \cdot e^{0.4t}$ .
To solve for $t$ , we can first divide both sides by 20: $\frac{40}{20} = \frac{20 \cdot e^{0.4t}}{20}$ which simplifies to $2 = e^{0.4t}$ .

STEP 9

Now, we can take the natural logarithm (ln) of both sides: $\ln(2) = \ln(e^{0.4t})$ .
This simplifies to $\ln(2) = 0.4t$ .

STEP 10

Finally, we can divide both sides by 0.4 to isolate $t$ : $t = \frac{\ln(2)}{0.4} \approx \frac{0.6931}{0.4} \approx 1.7328$ .
Rounding to the nearest tenth gives us $t \approx \textbf{1.7}$ days.

STEP 11

(a.) After 30 days, the population is 3,255,095.828 bacteria. (b.) It takes approximately 1.7 days for the population to double.

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Studdy Solution

STEP 1

What is this asking? We've got some bacteria multiplying like crazy, and we need to figure out how many there are after 30 days and how long it takes for their population to double. Watch out! Don't mix up the population itself with the *rate* at which it's growing!

STEP 2

1. Set up the equation2. Calculate the population after 30 days3. Calculate the time to double

STEP 3

STEP 4

In our case, the **initial population** P0P_0P0​ is **20**, and the **growth rate** kkk is **0.4**.So, our equation becomes P(t)=20⋅e0.4tP(t) = 20 \cdot e^{0.4t}P(t)=20⋅e0.4t.Awesome!

STEP 5

We want to find the population after 30 days, so we need to plug in t=30t = \textbf{30}t=30 into our equation: P(30)=20⋅e0.4⋅30P(30) = 20 \cdot e^{0.4 \cdot \textbf{30}}P(30)=20⋅e0.4⋅30.

STEP 6

Now, let's calculate: P(30)=20⋅e12P(30) = 20 \cdot e^{12}P(30)=20⋅e12.This gives us P(30)≈20⋅162754.7914=3255095.828P(30) \approx 20 \cdot 162754.7914 = 3255095.828P(30)≈20⋅162754.7914=3255095.828.So, after 30 days, there will be approximately **3,255,095.828** bacteria!

STEP 7

We want to find the time it takes for the population to double.That means we're looking for the time ttt when P(t)=2⋅P0P(t) = 2 \cdot P_0P(t)=2⋅P0​, which is 2⋅20=402 \cdot 20 = 402⋅20=40.

STEP 8

Let's set up the equation: 40=20⋅e0.4t40 = 20 \cdot e^{0.4t}40=20⋅e0.4t.To solve for ttt, we can first divide both sides by **20**: 4020=20⋅e0.4t20\frac{40}{20} = \frac{20 \cdot e^{0.4t}}{20}2040​=2020⋅e0.4t​ which simplifies to 2=e0.4t2 = e^{0.4t}2=e0.4t.

STEP 9

Now, we can take the natural logarithm (ln) of both sides: ln⁡(2)=ln⁡(e0.4t)\ln(2) = \ln(e^{0.4t})ln(2)=ln(e0.4t).This simplifies to ln⁡(2)=0.4t\ln(2) = 0.4tln(2)=0.4t.

STEP 10

Finally, we can divide both sides by **0.4** to isolate ttt: t=ln⁡(2)0.4≈0.69310.4≈1.7328t = \frac{\ln(2)}{0.4} \approx \frac{0.6931}{0.4} \approx 1.7328t=0.4ln(2)​≈0.40.6931​≈1.7328.Rounding to the nearest tenth gives us t≈1.7t \approx \textbf{1.7}t≈1.7 days.

STEP 11

(a.) After 30 days, the population is 3,255,095.828 bacteria. (b.) It takes approximately 1.7 days for the population to double.

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What is this asking? We've got some bacteria multiplying like crazy, and we need to figure out how many there are after 30 days and how long it takes for their population to double. Watch out! Don't mix up the population itself with the rate at which it's growing!

1. Set up the equation
2. Calculate the population after 30 days
3. Calculate the time to double

In our case, the initial population $P_0$ is 20, and the growth rate $k$ is 0.4.
So, our equation becomes $P(t) = 20 \cdot e^{0.4t}$ .
Awesome!

We want to find the population after 30 days, so we need to plug in $t = \textbf{30}$ into our equation: $P(30) = 20 \cdot e^{0.4 \cdot \textbf{30}}$ .

Now, let's calculate: $P(30) = 20 \cdot e^{12}$ .
This gives us $P(30) \approx 20 \cdot 162754.7914 = 3255095.828$ .
So, after 30 days, there will be approximately 3,255,095.828 bacteria!

We want to find the time it takes for the population to double.
That means we're looking for the time $t$ when $P(t) = 2 \cdot P_0$ , which is $2 \cdot 20 = 40$ .

Let's set up the equation: $40 = 20 \cdot e^{0.4t}$ .
To solve for $t$ , we can first divide both sides by 20: $\frac{40}{20} = \frac{20 \cdot e^{0.4t}}{20}$ which simplifies to $2 = e^{0.4t}$ .

Now, we can take the natural logarithm (ln) of both sides: $\ln(2) = \ln(e^{0.4t})$ .
This simplifies to $\ln(2) = 0.4t$ .

Finally, we can divide both sides by 0.4 to isolate $t$ : $t = \frac{\ln(2)}{0.4} \approx \frac{0.6931}{0.4} \approx 1.7328$ .
Rounding to the nearest tenth gives us $t \approx \textbf{1.7}$ days.