Here are the answers from the document you provided:

1. Power Gain (G = 8) in decibels (dB): $\text{Power Gain (dB)} = 10 \log_{10}(8) \approx 9.03 \, \text{dB}$

2. Voltage Gain (G = 4) in decibels (dB): $\text{Voltage Gain (dB)} = 20 \log_{10}(4) \approx 12.04 \, \text{dB}$

3. Converting power and voltage levels to dBW, dBm, and dBV:

   • If $P = 40 \, W$, then $P = 16.02 \, \text{dBW}$.
   • If $P = 240 \, mW$, then $P = 23.80 \, \text{dBm}$.
   • If $V = 120 \, V$, then $V = 41.58 \, \text{dBV}$.
   • If $P = 60 \, \text{dBW}$, then $P = 90 \, \text{dBm}$.

Let me know if you need further details or clarification.

Related Questions:

1. How is the logarithmic formula for decibels derived from the basic principles of logarithms?
2. Why is there a difference between voltage and power gain formulas in dB (using 20 vs 10)?
3. How can you convert dBm back to watts?
4. What are common applications of dB in computing and communication systems?
5. Can you explain the relationship between power in dBW and dBm?

Tip: Always remember that converting between dBW and dBm involves a simple 30 dB offset.