Math  /  Algebra

QuestionThe number of bacteria in a culture is given by the function n(t)=860e0.18tn(t)=860 e^{0.18 t} where tt is measured in hours. (a) What is the continuous rate of growth of this bacterium population?
Your answer is \square percent (b) What is the initial population of the culture?
Your answer is \square (c) When will the culture contain 2115 bacteria?
Your answer is \square Round to the nearest hour.

Studdy Solution

STEP 1

What is this asking? We're given a formula that tells us how many bacteria there are at any given time, and we need to figure out the growth rate, the starting number of bacteria, and how long it takes to reach a specific number of bacteria. Watch out! The formula already has *e* in it, so we don't need to add another one!
Also, pay close attention to units – time is measured in hours here.

STEP 2

1. Find the Growth Rate
2. Calculate the Initial Population
3. Determine the Time to Reach 2115 Bacteria

STEP 3

The formula n(t)=860e0.18tn(t) = 860e^{0.18t} is already in the form we need!
The **continuous growth rate** is the coefficient of the tt variable in the exponent.
In our case, that's **0.18**.

STEP 4

To express this as a percentage, we **multiply by 100**: 0.18100=18%0.18 \cdot 100 = 18\%.
So, the continuous growth rate is **18%** per hour.

STEP 5

The **initial population** is the number of bacteria at the very beginning, when t=0t = 0.
Let's plug that into our formula: n(0)=860e0.180n(0) = 860e^{0.18 \cdot 0}.

STEP 6

Since any number raised to the power of zero is one, we have e0.180=e0=1e^{0.18 \cdot 0} = e^0 = 1.
Therefore, n(0)=8601=860n(0) = 860 \cdot 1 = 860.
The initial population is **860** bacteria.

STEP 7

We want to find the time tt when n(t)=2115n(t) = 2115.
Let's set up the equation: 2115=860e0.18t2115 = 860e^{0.18t}.

STEP 8

To **isolate the exponential term**, we'll **divide both sides** by 860: 2115860=e0.18t\frac{2115}{860} = e^{0.18t}.
This simplifies to approximately 2.4593=e0.18t2.4593 = e^{0.18t}.

STEP 9

Now, we'll take the **natural logarithm (ln)** of both sides to get the tt out of the exponent: ln(2.4593)=ln(e0.18t)\ln(2.4593) = \ln(e^{0.18t}).

STEP 10

Using the property ln(ex)=x\ln(e^x) = x, we get ln(2.4593)=0.18t\ln(2.4593) = 0.18t.
Calculating the natural log, we have 0.89990.18t0.8999 \approx 0.18t.

STEP 11

Finally, we **divide both sides by 0.18** to solve for tt: t0.89990.184.999t \approx \frac{0.8999}{0.18} \approx 4.999.
Rounding to the nearest hour, we get t5t \approx 5 hours.

STEP 12

(a) The continuous growth rate is **18%**. (b) The initial population is **860** bacteria. (c) The culture will contain 2115 bacteria in approximately **5** hours.

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