Dude, you are a RIOT! Your units are all screwed up. Ask anybody who's taken basic algebra. This is an exponential function with a range on the interval [0, 1]. Try the units again, you'll find the attenuation WAY more than what you're expecting here. LMAO!!!!
All right rocketman,
From the sited reference, page 4:
P(R)/P(0) = e^(-alpha * R)
Where P(R) is power at distance R and P(0) is the power at the origin.
alpha is typically 0.1 (0.43dB/km) to 1.0 (4.3dB/km) at 1.0km. Hence the referenced formula is in 1.0km distance units which is easily proven by:
e^(-.1 * 1) = 0.90484 which 10 * log(.90484) = 0.43dB/km as sited in the reference.
5 miles = 8.0km or 8 times the formula's 1.0km distance unit.
I used an alpha of 0.115 (0.5dB/km), which is slightly worse than the clear sky number in the reference.
Therefore e^(-0.115 * 8.0) = 0.3985 which is the power ratio of the source to the destination at 8.0km.
I started with a power of 50 watts at my source.
0.3985 * 50 = 19.9.
Therefore a laser passing through 5 miles of clear sky conditions would have a power level of 19.9 watts at its destination as sited in my earlier example.
In engineering decibles are often used to calculate link loss among other things.
The easier way to calculate the above is this way.
As previously stated, an alpha of 0.115 is equivelent to 0.5dB/km of attenuation which was also sited in my first example.
0.5dB * 8 (for 8.0km) = 4dB of attenuation over 8.0km.
4dB is a power ratio of: 10^(-4/10) = 0.3981.
The same number as above (without rounding errors).
You lose again.