Expressing Gain in Decibels?
It's just a convention. Decibels always refer to power, and power is proportional to voltage squared and current squared. The math works like this:


$$\begin{align}
A_{v,dB} &amp;= 10\cdot\log \frac{V_o^2}{V_i^2} \\
         &amp;= 10\cdot\log \left(\frac{V_o}{V_i}\right)^2 \\
         &amp;= 10\cdot 2 \cdot\log \frac{V_o}{V_i}
\end{align} $$

EDIT: Power is proportional to voltage/current squared for all linear circuits. In AC circuit analysis, voltage, current, and power all become phasors. In nonlinear circuits, power may not be proportional to voltage/current squared, but the convention is still used for decibels. It even holds in an abstract field like signal processing, where the "power" of a signal is defined to be the average of the amplitude squared.


Decibels are used to give the magnitude of the gain. If you want to talk about power loss, you use negative decibels, which represents a fractional gain. For example:


$$A_p = 0.01$$
$$A_{p,dB} = 10\cdot\log 0.01 = -20\:\mathrm{dB}$$

You can also have an actual negative gain, like what you get from an inverting amplifier. The negative there is described as a phase shift. To fully describe an amplifier, you usually need both the magnitude and the phase of the gain. (This might be a bit advanced for you, but you could try looking up Bode Plots to see how this is used in real life.) Anyway, here's how to describe an inverting amplifier with a voltage gain of -2.4:

$$A_v = -2.4$$
$$|A_v| = 20\cdot\log 2.4 \approx 7.6\:\mathrm{dB}$$
$$\angle A_v = 180^\circ$$