Measuring LM7805-regulated voltage of battery

Im powering a nano and servo from a 7.4V battery pack. So Ive used an LM7805 to trim the voltage down to 5 for the nano and servo.

The thing is that part of the project also uses an LCD to display the voltage (as a proxy to SOC for my battery). So originally the LCD will display the 5V but Im wondering how that will vary as energy in the pack is consumed and voltage drops. Is it linear?

So will a drop from 5->4.7V (6% drop) = 6% * 7.4 = a 0.44 drop, resulting in 6.96V?

Is it linear?

No.

The actual curve depends on the battery chemistry.

As you have a 7805 regulator the voltage should stay at 5V until the battery voltage falls below the 7805 dropout voltage which is somewhere around 7 to 7.5V. After that the 5V will drop but exactly how much is not linear and not easy to calculate.

Steve