The ADF9010 baseband input impedance?

The input impedance of each pin is 250 Ohm, that means the differential impedance is 500Ohm: This is correct.

I  am doing the RF chain caculation. ADL5382--ADF9010.  I use ADL5382 as demodulator, the output I/Q differential impedance is 50 ohm, the voltagage gain is 4dB, however the input impedance of ADF9010 is 500 ohm(differential impedance), that means the power gain is Pgain=Vgain+10LOG(Rin/Rout), that induce 6dB loss. 

If i use 1:10 balun, there is no power loss?

Is this the right method to deal with power loss?

Thank you.

Another question, The ADL5382 voltage gain is about 3.9dB, the power gain is 3.9+10*LOG(Rin/RL), if the next stage is ADF9010, because the input differential  impedance is 500 ohm（24dB gain setting）。3.9-10.4=-6.5dB. The power loss is 6.5dB。we have to use impedance transformer in order to make there is no power loss？