A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is -5 dB per km and cable length is 20 km. The power received at receiver is ${10}^{-x}W$. The value of x is____. 

$\left[{\text{Gain in dB=10log}}_{10}\left(\frac{P_0}{P_i}\right)\right]$

Step 1. Given data

Power of signal transmitted

$P_i=0.1Kw=100w$

Total length of path $=20km$

Rate of attenuation $=-5dB/Km$

Step 2. Calculating the value of x

Total loss suffered $=-5\times 20=-100dB$

We know that Gain is given by

Gain in$dB=10{\log }_{10}\left(P_0/P_i\right)$

-$100=10{\log }_{10}\left(P_0/P_i\right)$

$\begin{array}{l}{\log }_{10}\left(P_i/P_0\right)=10\\ {\log }_{10}\left(P_i/P_0\right)={\log }_{10}{10}^{10}\\ \frac{100}{P_0}={10}^{10}\\ P_0=\frac{1}{{10}^8}={10}^{-8}\end{array}$