Reading through an RF power amplifier datasheet, I found the sheet only referenced dBm output power, rather than watts.
What is dBm?
dBm stands for decibels relative to one milliwatt. Decibels represent multiplicative factors, or ratios; by establishing a specific reference level they can instead be used as absolute values: 0 dBm is 1 milliwatt, 3 dBm is approximately 2 milliwatts, etc.
How do I convert it to watts?
Convert the decibel value to a scale factor and multiply by one milliwatt. That is,
$$x_{\mathrm{mW}} = 10^{x_{\mathrm{dBm}}/10} \cdot 1 \,\mathrm{mW}$$
$$x_{\mathrm{W}} = 10^{x_{\mathrm{dBm}}/10} \cdot 0.001 \,\mathrm{W}$$
For example, the datasheet you link mentions a value of
$$17 \,\mathrm{dBm} = 10^{17/10} \,\mathrm{mW} \approx 50.1 \,\mathrm{mW}$$
Why, or when, would you use dBm to specify power output rather than watts? Are there specific problems or equations that are easier to deal with in dBm vs watts?
Gain and loss in all stages of an RF system (feed line, filters, amplifiers) is multiplicative (if it were not, that would be nonlinearity), and therefore is typically written in dB so that the total gain or loss may be computed by adding, rather than multiplying, all the individual values together.
If you add a value in dB to a value in dBm, the result is in dBm. (Adding two dBm values is not usually meaningful since it would correspond to power squared.)
Are there specific problems or equations that are easier to deal with in dBm vs watts?
Decibel units, dBm being an example of such, provide a more intuitive measure of some property that responds logarithmically, like power frequently does.
Consider, if you are transmitting now with 1W, and you add 1W more, you have doubled your transmit power. That's a big difference.
If you are transmitting with 100W, and you add 1W more, your transmit power is 101W. This is only 1% more power: hardly a relevant change.
Decibels account for this. From 1W to 2W is a +3dB change. From 100W to 101W is a +0.043dB change. If you were to increase +3dB from 100W, the result would be 200W, which is the same degree of improvement as 1W to 2W is.
Power dB = 10 log10 (ratio)
Power in dBm helps in quickly calculating the power at Rx end. Say a launch power of +13 dBm (=10^(1.3) mW = 20 mW) - with link loss of 6 dB - would give 13-6 = 7 dBm at the Rx end. We can verify.
6dB = 10^(.6) = 4.
-6dB = 1/4.
So output power = 20 mW * 1/4 = 5 mW.
10*log10 (5) = 7 dBm.
So, it checks out.
As suggested when launch power is in dBm - one can just subtract the losses in dB to get the dBm power received at the Rx end.
Decibels turn multiplication/division problems into addition/subtraction problems and keep the numbers more human-friendly.
A key thing to remember is that typically in radio, only a tiny fraction of the transmit power actually makes it to the receiver. So in a radio system we typically deal with signals that are quite large and signals that are very small. Furthermore, the uncertainty is also typically large, a factor of many times difference in the result can easily be within the margin of error.
Pure decibels are a logarithmic scale that can be used to compare the level of two signals, such as the input and output of an amplifier, or the power at the receive antenna compared to the power at the transmit antenna. There are two related conventions, which are equivalent if the system impedance is constant, but differ if the two signals are driven into different impedances.
For signals measured in terms of power, we take the base 10 log of the power ratio and multiply it by 10. 10dB represents 10 times the power. 20dB represents 100 times the power, 30dB 1000 times the power and so-on. 1dB represents about 1.25 times the power, 3dB represents about twice the power.
For signals measured in terms of voltage, we take the base 10 log of the voltage ratio and multiply it by 20. 20dB represents 10 times the voltage, 40dB represents 100 times the voltage 60dB 1000 times the voltage and so-on.
That's fine for comparing the power (or voltage) of signals, but what if we want to talk about a signal in isolation? Well then we speak of decibels relative to a reference.
In radio dBm is relative to 1 milliwatt (the situation in audio is a little more complex). Therefore 0 dBm is one milliwatt. 30dBm is 1 watt, 60dBm is one kilowatt. 90dBm is one megawatt. Going the other way -30dBm is one microwatt, and -60dBm is one nanowatt.
So if we look at say frizz transmission equation for radio transmission in free space (sadly we don't always have the luxury of free space, but that's another subject) specified in linear units with gains measured relative to an isotropic antenna.
$$P_r = P_t G_t G_r \left( \frac{\lambda}{4 \pi d} \right)^2$$
Lets say $\lambda$ is around 10 centimetres and d is around say 10 meters, then $\left( \frac{\lambda}{4 \pi d} \right)^2$ is around $6 \times 10^{-7}$. It's certainly not impossible to work with numbers like that, but it's inconvenient and error prone.
Now suppose instead of linear units, we use decibel units for the transmit and receive power, and we use decibels relative to an isotropic antenna for the antenna gains. We keep the linear units for distance and wavelength. Our equation becomes.
$$P_r = P_t + G_t + G_r + 20\log_{10}\left( \frac{\lambda}{4 \pi d} \right)$$
Now, instead of dealing with multiplication and a tiny value for $\left( \frac{\lambda}{4 \pi d} \right)^2$ we are dealing with addition/subtraction and a far more human-scale value of about minus 62 for $20\log_{10}\left( \frac{\lambda}{4 \pi d} \right)$.
dBm is an incomplete term
You also need the missing parameter which the datasheet writers assume you know: the impedance. For audio / telegraphy this is 600 ohms but for RF this is usually 50 ohms.
