One reason is that the exponential expression gives the same result wherever you start. If going from 0m to 1m gives a ratio of 0.1, then going from 1m to 2 m will give 0.1, and so will going from 2m to 3m and 3m to 4m. Attenuation from 0 to 2m will be 0.01. It all the ratios for the shorter distances multiplied together. This is a property of the exponential function that your other suggestions don't have. Decibels is the log of the ratio, and multiplication of the ratios is described by addition of the logarithms, which is why you add dB for each part of a path.

For the other two of your suggestions the power ratio depends on where you start. Going from 0m to 1m gives an infinite power ratio, from 1m to 2m gives 1/2 or 1/4, and from 2m to 3m gives 2/3 or 4/9. This is not how attenuation due to loss in materials is observed to behave.

Something that does behave in that way is "space attenuation" which is the reduction in amplitude seen when a lossless wave travels outwards cylindrically or spherically. In these cases there is a zero point to work from, and it does follow a 1/r or 1/r2 law. The observed attenuation rate in dB/m changes with distance from the centre, and at long ranges approaches zero, as the wave itself also becomes similar to a plane wave over widths small compared to the range.

It's amazing, how they seam to find their fuel everywhere, and what if they make a journey into space visiting the Musk car, bringing around latest technology (but I don't believe it, too large for them). Spatially, they may curve into themselves, and releasing like a torsion spring, to achieve kinetic energy, first time, and then only remember where the receiver is. No use with expressions like fuel and what's left...?

I think that it is because they are different things. The first one corresponds to the propagation of a wave (some perturbation of a field). The second one is the law of decreasing of an intensity of a field with the distance from the source.

For signal propagation in a cable you have exponential decay..for propagation in free space its proportional 1/r^2..thats why fress space is often better.(antenna transmission)

As I remember, there is a solution method ; faltung with the fundamental solution, and then the denominator could be (x-a)^2, or (x-a)

Pierre Bouguer was probably first to ask this question and his answer was put by Heinrich Lambert into math. The corresponding differential equation leads to an exponential decay in an absorbing medium. Although this was before electromagnetics, what is valid for irradiance is also valid for the electric field...

Dr Mayerhöfer,

Interesting, but did not know; it would be a first order equation then.

There is also the Brownian motion. It looks in agreement with one example in Lovelock and Rund, and their method. I used it on a soliton equation and arrived in a new equation with a solution 1/(x-ct)^2. Below

Article Models for Elasto-electricity and PhotoVoltaic in a Micro Power Plant

I agreed with Mr. Thomas Mayerhöfer, In additional, I think that the exponential attenuation model, with

I(x) = I_0 e^{-alpha*x}

provides a physically meaningful representation of how intensity diminishes as a wave travel through a medium (from source to infinity).

In contrast, an inverse power law such as

I(x) = I_0/x^n

would imply infinite intensity at the source (lim(x->0) = infinity), which is not physically acceptable.

Adam Jouamaa,

It depends on in what medium. Traditionally, when regarding safety, such solutions were not disregarded. If it comes to gaining power, maybe you can't rely on it, unless a material responds with interaction. Maybe I don't understand the question. Kindest regards

Shalabh Kumar Mishra Lena J-T Strömberg Thomas Mayerhöfer Adam Jouamaa Fritz Caspers Luis Gaete-Garretón Malcolm White

Electromagnetic waves attenuate exponentially in a medium because the energy loss due to absorption and scattering is proportional to the wave's intensity at each point. This results in an exponential decay, E(x)=E0e−γxE(x) = E_0 e^{-\gamma x}E(x)=E0​e−γx. Inverse or cubic attenuation would describe different physical scenarios and do not apply to this type of energy loss.

Mr Mishra, and others.

We know that waves travels on wave-guards for example around the Earth while communicating like this. Is the question about safety, or more general? I imagine you are the Mayor in the city and should decide where to put wifi and equipments to not over-expose. For light, we know that it turns (lateral light), and the plants grow at the shadow side, while on the other side develop other things. Behind; the shadow, is actually due to light, so it's energetic for some discussions as well. I worked at a ship, before there were internet, so I remember some of the security regulations for EM. Sun overexpose, and the magnetic field shields the Earth, according to science.