radar attenuation calculations
I wonder exactly why you care about the radar range equation, given you aren't an engineer. I'd love to hear about that.

If you read the text you will see exp used. The exp function is the value \$e\$ (as you point out, a number close to 2.718) raised to the given power. For example, \$e^x=\operatorname{exp}\left(x\right)\$. So, yes, in the equation that is what \$e\$ is.

In your radar range example, the idea they are talking about isn't complicated.

Imagine the radar emits rays that proceed outward and are partially absorbed as they pass through the air. We can imagine that a certain percentage remains after traveling some given distance. When it travels further along for the same distance, there will be a percentage of the original percentage. And so on.

This is a multiplicative process (repeated multiplication.) Such processes are easily represented by using power equations (like \$a^b\$.) The above description can be turned into a mathematical expression. Let's say that the remaining percentage of a ray, after traveling a certain distance, is called \$\gamma\$ (we can call it anything.) Then after twice this distance we'd have \$\gamma^2\$ left over. And after three times the distance we'd have \$\gamma^3\$ remaining.

Assuming that \$R\$ is the distance (specified in those certain units), then we can say that what remains of each ray is \$\propto \gamma^R\$ (reads: proportional to \$\gamma^R\$.) However, since your equation deals with both the outgoing and return paths where the transmitter and receiver are at the same location, the total path length is twice the maximum range. So this means \$\propto \gamma^{2\cdot R}\$ (to account for the total path being twice the maximum range.)

This can be transformed:

$$\begin{align*}
\propto &amp;\quad \gamma^{2\cdot R}\\\\
&amp;\quad e^{\operatorname{ln}\left(\gamma^{2\cdot R}\right)}\\\\
&amp;\quad e^{2\cdot R\:\operatorname{ln}\gamma}
\end{align*}$$

Since \$\gamma\$ is less than 1, \$\operatorname{ln}\gamma\$ is negative. If we treat \$\alpha=-\operatorname{ln}\gamma\$ (so that \$\alpha\$ is positive [or zero]) then the resulting factor is \$\propto\:e^{-2\cdot \alpha\cdot R}\$.

Since decibels for power are based on \$10\cdot\operatorname{log_{10}}\left(x\right)\$ (and not the attenuation factor, directly), you can re-express the prior ratio in decibels as:

$$\begin{align*}
&amp;{10\cdot\operatorname{log_{10}}\left(e^{-2\cdot \alpha\cdot R}\right)}\\\\
&amp;{-2\cdot \alpha\cdot R\cdot 10\cdot\operatorname{log_{10}}\left(e\right)}\\\\
&amp;{-2\cdot \alpha\cdot R\cdot 4.34294482}\\\\
&amp;{-2\cdot \left(4.34294482\cdot \alpha\right)\cdot R}
\end{align*}$$

Finally, yes. You do need to solve for \$R_\text{MAX}\$ and to do that you will need the LambertW (product-log) function. It's not at all hard to apply here. If you need help with it, I can walk you through the few lines of algebraic manipulation (it's not much work at all.)



Let's lump all the factors into a single constant, \$k_0\$, such that:

$$R_\text{MAX}^4=k_0\cdot e^{-2\,\alpha\, R_\text{MAX}}$$

(I'm sure you can see what \$k_0\$ is without forcing me to write out all those factors in the photocopied equation you have in your question.)

Then just follow the steps below:

$$\begin{align*}
R_\text{MAX}^4&amp;=k_0\cdot e^{-2\,\alpha\, R_\text{MAX}}\\\\
\sqrt[\leftroot{0}\uproot{2}4]{R_\text{MAX}^4}&amp;=\sqrt[\leftroot{0}\uproot{2}4]{k_0\cdot e^{-2\,\alpha\, R_\text{MAX}}}\\\\
R_\text{MAX}&amp;=\sqrt[\leftroot{0}\uproot{2}4]{k_0}\cdot e^{-\frac12\,\alpha\, R_\text{MAX}}\\\\
R_\text{MAX}\cdot e^{\frac12\,\alpha\, R_\text{MAX}}&amp;=\sqrt[\leftroot{0}\uproot{2}4]{k_0}\\\\
\frac12\,\alpha\,R_\text{MAX}\cdot e^{\frac12\,\alpha\, R_\text{MAX}}&amp;=\frac12\,\alpha\,\sqrt[\leftroot{0}\uproot{2}4]{k_0}
\end{align*}$$

At this point, we just recall that if \$u\,e^u=v\$, then \$u=\operatorname{LambertW}\left(v\right)\$. So, we set \$u=\frac12\,\alpha\,R_\text{MAX}\$ and then:

$$\begin{align*}
\left(\frac12\,\alpha\,R_\text{MAX}\right)\, e^{^{\left(\frac12\,\alpha\,R_\text{MAX}\right)}}&amp;=\frac12\,\alpha\,\sqrt[\leftroot{0}\uproot{2}4]{k_0}\\\\
u\, e^{u}&amp;=\frac12\,\alpha\,\sqrt[\leftroot{0}\uproot{2}4]{k_0}\\\\
&amp;\therefore\\\\
u&amp;=\operatorname{LambertW}\left(\frac12\,\alpha\,\sqrt[\leftroot{0}\uproot{2}4]{k_0}\right)\\\\
\frac12\,\alpha\,R_\text{MAX}&amp;=\operatorname{LambertW}\left(\frac12\,\alpha\,\sqrt[\leftroot{0}\uproot{2}4]{k_0}\right)\\\\
R_\text{MAX}&amp;=\frac2{\alpha}\,\operatorname{LambertW}\left(\frac{\alpha}2\,\sqrt[\leftroot{0}\uproot{2}4]{k_0}\right)
\end{align*}$$

It's that simple. (Well, you get to hand-write out \$k_0\$ now. I don't want to have to write all that stuff, myself. But you know how to do that last step, I'm sure.)