Derivation the dB version of the Path Loss Equation for Free Space.

For free propagation waves in radio channel, the path loss model is

$$L_0 = { \left ( \frac{4 ~ \pi ~ d}{\lambda} \right ) }^2$$ (1)

Where $\lambda = c / f_c$ , so

$$L_0 = { \left ( \frac{4 ~ \pi ~ d ~ f_{c}}{c} \right ) }^2$$ (2)

For d in meters , f_c in GHz and $c = 3 \times 10^8$ meters / second,

$$L_0 = {\left ( \frac{4 ~ \pi ~ d ~ f_c \times 1 \times 10^9}{3 \times 10^8} \right ) }^2$$ (3)

By taking $\log_{10} \left ( \log \right )$ of both sides of equation $\left ( 3 \right )$ to obtain the dB version

$$10 \log { \left ( \frac{4 ~ \pi ~ d ~ f_c \times 1 \times 10^9 }{3 \times 10^8 } \right ) }^2$$ L₀ =

$$20 \log \left ( \frac { 4 \pi \times 10 }{3} \right ) + 20 \log \left ({f_c} \right ) + 20 \log \left ({d} \right )$$ =

$$32.4 + 20 \log \left ({f_c} \right ) + 20 \log \left ({d} \right )$$ = (4)