*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 decibels (dB) and power changes, we use the formula for calculating decibels in terms of power:

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

where $P_2$ is the final power and $P_1$ is the initial power.

1. **Correct Answer: Approximately 3 dB**

   - A factor of two increase in power means $P_2 = 2 \cdot P_1$. Plugging this into the formula gives:
     $$
     \text{dB} = 10 \cdot \log_{10}(2) \approx 3 \, \text{dB}
     $$
   - Similarly, a factor of two decrease in power means $P_2 = \frac{1}{2} \cdot P_1$. Plugging this into the formula gives:
     $$
     \text{dB} = 10 \cdot \log_{10}\left(\frac{1}{2}\right) \approx -3 \, \text{dB}
     $$
   - Therefore, a factor of two change in power (either increase or decrease) is approximately 3 dB.

2. **Why the Other Choices are Incorrect:**

   - **Choice A: Approximately 2 dB**  
     - 2 dB does not correspond to a factor of two change in power. It is too small to represent this change.
   
   - **Choice C: Approximately 6 dB**  
     - 6 dB represents a factor of four change in power, not two. This is because:
       $$
       10 \cdot \log_{10}(4) \approx 6 \, \text{dB}
       $$
   
   - **Choice D: Approximately 9 dB**  
     - 9 dB represents a factor of eight change in power, not two. This is because:
       $$
       10 \cdot \log_{10}(8) \approx 9 \, \text{dB}
       $$

In summary, a factor of two increase or decrease in power is approximately 3 dB, making choice B the correct answer.