Let's solve this step by step.

We are given the equation for the number of bacteria:

$P(h) = 1600e^{0.18h}$

We want to determine the time $h$ when $P(h) = 2200$. Start by substituting 2200 into the equation:

$2200 = 1600e^{0.18h}$

Step 1: Divide both sides by 1600

$\frac{2200}{1600} = e^{0.18h}$

$1.375 = e^{0.18h}$

Step 2: Take the natural logarithm (ln) of both sides

$\ln(1.375) = 0.18h$

Step 3: Solve for $h$

$h = \frac{\ln(1.375)}{0.18}$

Calculate $\ln(1.375)$ and divide by 0.18:

$\ln(1.375) \approx 0.31845$

$h = \frac{0.31845}{0.18} \approx 1.7692 \, \text{hours}$

Step 4: Round to the nearest tenth

$h \approx 1.8 \, \text{hours}$

Final Answer:

It will take approximately 1.8 hours for the bacteria population to reach 2200.

Let me know if you'd like a detailed breakdown of any step or further explanation! Here are some related questions to expand your understanding:

1. What is the doubling time of this bacterial population?
2. How would the time change if the initial population was 2000 instead of 1600?
3. What happens to $h$ if the growth rate constant changes from 0.18 to 0.2?
4. How can we rewrite the equation using logarithmic properties for ease of computation?
5. What is the population after 3 hours?

Tip: Always isolate the exponential term first when solving these types of equations!