Four years from now, the population of a colony of bees will reach `1.6xx10^(8)`. If the population of the colony doubles every 2 years, what was the population 4 years ago?

Answer

To find the population of the colony of bees four years ago, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the future population**: We know that the population of the colony of bees will reach \(1.6 \times 10^8\) in four years.

2. **Determine the time interval**: The time from four years ago to four years from now is a total of 8 years (4 years ago + 4 years from now).

3. **Calculate the number of doublings**: Since the population doubles every 2 years, we can find out how many times it has doubled in 8 years:
\[ \text{Number of doublings} = \frac{8 \text{ years}}{2 \text{ years/doubling}} = 4 \text{ doublings} \]

4. **Set up the equation**: Let the population four years ago be \(X\). After 4 doublings, the population will be:
\[ X \times 2^4 = 1.6 \times 10^8 \]

5. **Calculate \(2^4\)**: We know that:
\[ 2^4 = 16 \]
Therefore, we can rewrite the equation as:
\[ X \times 16 = 1.6 \times 10^8 \]

6. **Solve for \(X\)**: To find \(X\), we can rearrange the equation:
\[ X = \frac{1.6 \times 10^8}{16} \]

7. **Simplify the right side**: We can simplify \(1.6\) as follows:
\[ 1.6 = \frac{16}{10} \]
Thus, we can rewrite the equation:
\[ X = \frac{16}{10} \times \frac{10^8}{16} \]
The \(16\) cancels out:
\[ X = \frac{10^8}{10} = 10^7 \]

8. **Final answer**: Therefore, the population of the bees four years ago was:
\[ X = 1 \times 10^7 \]

### Conclusion:
The population of the colony of bees four years ago was \(1 \times 10^7\).