To solve this problem, we use the formula for exponential growth:

$P(t) = P_0 e^{kt}$

Where:

• $P(t)$ is the population at time $t$,
• $P_0$ is the initial population,
• $k$ is the growth rate,
• $t$ is the time elapsed.

Step 1: Write two equations based on given data.

At $t = 3$, $P(3) = 10,000$: $10,000 = P_0 e^{3k}$

At $t = 10$, $P(10) = 90,000$: $90,000 = P_0 e^{10k}$

Step 2: Divide the two equations to eliminate $P_0$.

Dividing the second equation by the first gives: $\frac{90,000}{10,000} = \frac{P_0 e^{10k}}{P_0 e^{3k}}$

Simplify: $9 = e^{10k - 3k} \quad \Rightarrow \quad 9 = e^{7k}$

Take the natural logarithm on both sides: $\ln(9) = 7k \quad \Rightarrow \quad k = \frac{\ln(9)}{7}$

Calculate $k$: $k \approx \frac{2.197}{7} \approx 0.3139$

Step 3: Solve for $P_0$ using one of the original equations.

Substitute $k$ into the equation $10,000 = P_0 e^{3k}$: $10,000 = P_0 e^{3(0.3139)}$

Calculate $e^{3(0.3139)}$: $e^{3(0.3139)} \approx e^{0.9417} \approx 2.564$

Now solve for $P_0$: $P_0 = \frac{10,000}{2.564} \approx 3,900$

Final Answer:

The initial number of bacteria was approximately: $\boxed{3,900}$

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Would you like additional explanation or have further questions?

Related Questions:

1. How do exponential growth and decay differ in terms of equations and behavior?
2. Can this problem be solved using logarithmic properties without dividing equations?
3. What is the doubling time for the bacterial population in this scenario?
4. How does changing the growth rate affect the initial population calculation?
5. What real-world factors could make a bacterial growth model deviate from exponential growth?

Tip: Always check your growth rate calculation by substituting it back into the original equations!