How to solve this logarithm expression? (SOLVED)

[knight]

Can anyone help me to find Pr(dBm) using this expression? Pr(dBm) = Pt(dBm) + Gt(dBi) + Gr(dBi) + PL(dB).

Given: Pt = 37dBm, Gt = 1.8dBi, Gr = 1.85dBi and PL = -38.4dB.

The answer given is Pr = -29.75dBm but I'm getting +32.25dBm. Thanks.

[radiolistener]

sorry, but this is bullshit. There is no logarithm expression, just a school arithmetic...

Pr = 37 dBm + 1.5 dBi + 1.85 dBi - 38.4 dBi = 1.95 dBm

[knight]

sorry, but this is bullshit. There is no logarithm expression, just a school arithmetic...

Pr = 37 dBm + 1.5 dBi + 1.85 dBi - 38.4 dBi = 1.95 dBm

Play it nicely bro. It is indeed a logarithm expression. According to your logic, 1dBm + 1dBm = 2dBm but that's incorrect. 

[radiolistener]

Play it nicely bro. It is indeed a logarithm expression. According to your logic, 1dBm + 1dBm = 2dBm but that's incorrect. 

No, your expression describes power of a single source with several gain described in dB or dBi units. dBi means signal gain relative to isotropic radiator.
If you want to sum two power sources the result depends on the source properties and measurement point.

For example, when sources are coherent and anti-phase, they cancel each other, so the sum of their power will be zero, in your case:
1 dBm + 1 dBm = -infinity dBm

When sources are coherent and in-phase the sum will be 20 * log10( 10^(P1/20) + 10^(P2/20), in your case:
1 dBm + 1 dBm = 7.02 dBm

When sources are incoherent the sum will be 10 * log10( 10^(P1/10) + 10^(P2/10) ), in your case:
1 dBm + 1 dBm = 3.51 dBm

So, there is no way to say what is the sum of two power sources when you don't know what is phase relation between these sources.