*Question G5B01:
What dB change represents a factor of two increase or decrease in power?*  

*Correct Answer is  B: Approximately 3 dB*  

### AI Response: (may contain errors)  

To understand the relationship between power changes and decibels (dB), we can use the following formula:

$$
dB = 10 \log_{10} \left( \frac{P_2}{P_1} \right)
$$

where:
- $P_1$ = initial power
- $P_2$ = final power

### Correct Answer: Approximately 3 dB

- A factor of two increase in power means $P_2 = 2P_1$. Plugging this into the formula gives:

$$
dB = 10 \log_{10} \left( \frac{2P_1}{P_1} \right) = 10 \log_{10}(2) \approx 10 \times 0.301 = 3.01 \text{ dB}
$$

- Therefore, a doubling of power corresponds to an increase of approximately 3 dB.

### Why the Other Choices are Incorrect:

- **A: Approximately 2 dB**
  - This is incorrect because a 2 dB change corresponds to a power ratio of about 1.58, not 2. 

- **C: Approximately 6 dB**
  - This is incorrect because a 6 dB change corresponds to a power ratio of 4 (since $10 \log_{10}(4) \approx 6$). This represents a fourfold increase, not a doubling.

- **D: Approximately 9 dB**
  - This is incorrect because a 9 dB change corresponds to a power ratio of 8 (since $10 \log_{10}(8) \approx 9$). This represents an eightfold increase, which is much greater than a doubling.

### Summary

- A doubling of power results in an increase of approximately 3 dB.
- Remember the key relationship: 
  - A change of +3 dB = doubling power
  - A change of -3 dB = halving power

This understanding is crucial for the Extra Class exam and practical applications in amateur radio.