We will solve each part of the problem step by step:

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a. How many bacteria were there initially?

The initial number of bacteria corresponds to $t = 0$.

The growth model is: $n(t) = 800(1.03)^t$

Substitute $t = 0$: $n(0) = 800(1.03)^0$ Since $(1.03)^0 = 1$: $n(0) = 800$

Answer: There were 800 bacteria initially.

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b. How many bacteria were there after 5 hours?

To find the population after 5 hours, substitute $t = 5$ into the growth model: $n(5) = 800(1.03)^5$

First, calculate $(1.03)^5$: $(1.03)^5 \approx 1.159274$

Now substitute: $n(5) = 800 \cdot 1.159274 \approx 927.419$

Answer: After 5 hours, there were approximately 927 bacteria.

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c. When does the bacteria population double?

To find when the population doubles, we set $n(t) = 2 \cdot 800 = 1600$ and solve for $t$: $1600 = 800(1.03)^t$

Divide both sides by 800: $2 = (1.03)^t$

Take the natural logarithm ($\ln$) of both sides: $\ln(2) = \ln((1.03)^t)$

Using the logarithmic property $\ln(a^b) = b \cdot \ln(a)$: $\ln(2) = t \cdot \ln(1.03)$

Solve for $t$: $t = \frac{\ln(2)}{\ln(1.03)}$

Calculate the logarithms: $\ln(2) \approx 0.693147, \quad \ln(1.03) \approx 0.029559$

Substitute: $t = \frac{0.693147}{0.029559} \approx 23.45$

Answer: The bacteria population doubles approximately after 23.45 hours.

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Summary of Results:

a. Initially, there were 800 bacteria.
b. After 5 hours, there were approximately 927 bacteria.
c. The population doubles after approximately 23.45 hours.

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Would you like a detailed explanation of any step, or have further questions?

Related Questions:

1. How does the growth rate ($1.03$) affect the doubling time?
2. What happens if the growth model is changed to $n(t) = 800(1.05)^t$?
3. How can you find the time when the bacteria population triples?
4. What is the population after 10 hours?
5. How does this model change if the initial population is 1000 instead of 800?

Tip:

Use logarithms whenever exponential equations are involved in finding unknown time or growth parameters.