Meaning of dB while dealing with amplifier(Op-Amp)
Here's some common dB values for voltage amplification:

20dB gain => \$10^{\left(\frac{20}{20}\right)}=10\$

6dB gain => \$10^{\left(\frac{6}{20}\right)}≈2\$

3dB gain => \$10^{\left(\frac{3}{20}\right)}≈\sqrt2\$

As you might have noticed, the way to convert voltage gain in dB to linear numbers is to simply use this equation:  \$f(x)=10^{\left(\frac{X}{20}\right)}\$

When you are talking about voltage gain you calculate the dB to \$20×\log\left(\frac{V_{out}}{V_{in}}\right)\$, when you are talking about power gain, then it's \$10×\log\left(\frac{P_{out}}{P_{in}}\right)\$. They are essentially the same thing, just keep your tongue in the correct mouth and everything will be easy peasy. The difference is that the voltage gain calculation assumes that R is the same which leads to the current, \$I\$ being the same, which is not always the case. So that's why the power equation is still there. 

And here's why! 

\$P=V×I\$ and \$I=\frac{V}{R}\$, if you replace the \$I\$ in the first equation you get \$P=\frac{V^2}{R}\$

So the original equation for power gain is \$10×\log\left(\frac{P_{out}}{P_{in}}\right)\$, which is the same as: 

$$10×\log\left(\frac{\frac{V_{out}^2}{R}}{\frac{V_{in}^2}{R}}\right)$$

The R cancel each other out. So you get: 

$$10×\log\left(\frac{V_{out}^2}{V_{in}^2}\right)$$

And that can be rewritten like this: 

$$10×\log\left(\left(\frac{V_{out}}{V_{in}}\right)^2\right)$$ 

And this is a logarithmic function, which means we can put the exponent in front of the logarithm. 

$$2×10×\log\left(\left(\frac{V_{out}}{V_{in}}\right)\right)$$ 

And that leaves us with the other equation, which is the same thing. 

$$20×\log\left(\left(\frac{V_{out}}{V_{in}}\right)\right)$$