I understand that you don't understand. The impedance of a speaker is basically the electrical resistance of a speaker. The higher the impedance (the higher the resistance) the HARDER it is to drive it. That's why head phones have higher impedance values (around 32-75 ohms ).

A subwoofer that have dual 4ohm coils have the same efficiently rating as their 2 ohm coiled variants. The only difference is how much power they suck out of an amplifier when they are hooked up. That's why you always match the amp with the speaker.

If two speakers receive the SAME wattage and the only thing different is their impedance value, then they will sound the same.

 

I think you meant to say the opposite. The higher the load the less hard the amp needs to work. That's why headphones can be driven by an opamp with a PSU the size of a thimble. Similarly, this is the reason why vacuum tube designs, low on current due to the nature of the beast, typically feature 16, 8, and 4 Ohm taps on their transformers.

1.You're talking about efficiency and sensitivity as if they're one and the same. They're not.
2.The power is not the only difference, as it doesn't exist in a vacuum. The distortion goes up, as does current draw, and the dampening halfs. We've talked about this. Please read the material, instead of trying to argue.

That's not necessarily true, assuming such a thing can exist. I mean, identical drivers posing alternate loads to an amplifier. I'm not talking about wiring options; it's about the drivers, themselves. Sonically, they may or may not sound the same, depending on implementation. Honestly, this is the audio equivalent of bench racing. OTOH, if you mean gain the same amplitude, well, that's not really possible even on a very simple mathematical level, again, all constraints staying the same. To wit:

As no amplifier is a perfect voltage source, its power does not double (perfectly) as the load halfs. Say the amp is rated at 100wpc @ 4 Ohms. That's 20 dBw (decibel watts). By the nature of the best, the same amp cannot reach 200wpc at 2 Ohms, unless its output at 4 ohms is actually greater than 100wpc, all things being equal. Remember, no such thing as a perfect voltage source. If it were a perfect voltage source, the amp would put out 200wpc at 2 Ohms, which translates to 23dBw. 20dBw @ 4 Ohms = 23dBw @ 2 Ohms. As we've already discussed, these values cannot be perfect, so the value at 4 Ohms will always be greater allowing for a more dBw, allowing for greater amplitude.

The above is said under the assumption that the drivers are identical. As such a thing cannot happen, the actual (audible) result may be a bit different than the one asserted above.

 

Let's put your statement to the test:

If I hook up a 75 ohm speaker to my amplifier will it be as loud as an exact replica (but) 4 ohm speaker? I mean, I'm sure if you adjust the gain and increase the input voltage (amplifiers have a limit to how many volts the input signal can be) it will be? right?

Yeah, I didn't think so.

 

For whatever reason, you seem to think that an amplifier having to work less hard somehow means louder, when no such correlation can be made in a vacuum, just as you assert that identical driver can exists whose loads differ by a factor of almost 20. I don't even know where to begin explaining this, if nothing else made a dent. Best of luck to you.

 

Correct.
But you need to consider that this is limited to the ideal case. In the real world, the voltage source used to set up the drive stage would see more losses at lower load - since current would increase all other parasitic circuit losses go up too.
Well, that's not entirely true either.... thinking about the drive stage itself, it's based on a varying linear regulator. Meaning is acts like a variable resistor in series with the load. The harder the output switch is turned on, the more current it can deliver - and sometimes, this can actually be done with less heat dissipated at the switch itself - but I'll stop here.