I haven't done this for a long time so check my answer.
Remember that the decibel, dB, is a logarithmic scale. Adding logs is the same as multiplication of the "un-logged" version of the numbers so that isn't going to work.
Instead we need to get the anti-log of each, sum them and get the log of the result.
$$
\begin{align}
P & = 10 log_{10}( 10^{\frac {-25}{10} } + 10^{\frac {-47}{10} } ) \\
& = 10 log_{10}( 0.003162 + 0.00001995 ) \\
& = 10 log_{10}(0.003182) \\
& = -24.975 \ \text {dBm}
\end{align}
$$
I'm expecting it to make a little bit more of a difference than that. Can anyone see an error?
The benefits of using dB are explained well in the Wikipedia decibel article:
Level values in decibels can be added instead of multiplying the underlying power values, which means that the overall gain of a multi-component system, such as a series of amplifier stages, can be calculated by summing the gains in decibels of the individual components, rather than multiply the amplification factors; that is, log(A × B × C) = log(A) + log(B) + log(C). Practically, this means that, armed only with the knowledge that 1 dB is approximately 26% power gain, 3 dB is approximately 2× power gain, and 10 dB is 10× power gain, it is possible to determine the power ratio of a system from the gain in dB with only simple addition and multiplication.