The number of bacteria $t$ days after the start of the experiment is given by the function:

$N(t) = k \cdot 3^{t/3}$

We are given that 9 days after the start of the experiment, the number of bacteria is 27,000. So, we can set up the equation:

$N(9) = k \cdot 3^{9/3} = 27,000$

Simplifying the exponent:

$N(9) = k \cdot 3^3 = k \cdot 27$

Now, we can solve for $k$:

$k \cdot 27 = 27,000$

$k = \frac{27,000}{27} = 1,000$

Now that we know $k = 1,000$, we can find the number of bacteria at the start of the experiment ($t = 0$):

$N(0) = 1,000 \cdot 3^{0/3} = 1,000 \cdot 3^0 = 1,000 \cdot 1 = 1,000$

So, there were 1,000 bacteria at the start of the experiment.

Would you like more details or have any questions?

Here are 5 related questions:

1. How would the number of bacteria change if $k$ were doubled?
2. What if the growth factor was $2$ instead of $3$?
3. How many bacteria would be present after 15 days?
4. What does the constant $k$ represent in this context?
5. How does the exponential growth rate compare to linear growth?

Tip: Understanding the base of the exponential function is crucial, as it determines how quickly the population grows over time.