It has been a very long time since I have done this kind of math. I have a tube type preamplifier for a friend. I want to make a solid-state equivalent just to test it against.
This amplifier has a gain of 39 BB, 42 DB, 45DB. I think without knowing that if I divide these numbers by 20 DB I will end up with a voltage gain, correct?
39DB=1.95 Gain
42DB= 2.1 gain
45 DB= 2.25 gain
is this correct or is my brain fart smellier than usual?

 

You were almost there.
Take dB value and divide by 20. Then raise 10 by that exponential, i.e. 10^(dB/20).

6dB = x2 gain
12dB = x4
18dB = x8
24dB = x16
30dB = x32

39dB = x89
42dB = x126
45dB = x178

dB = 20 x log(Vout/Vin)

https://www.allaboutcircuits.com/tools/decibel-calculator/

 

The division by 20 is for voltage gain or current gain. For power gain you would use a factor of 10 because power is proportional to voltage or current squared.

\( dB(\text{Power})\;=\;10 *log_{10}\left( \cfrac{P_{out}}{P_{in}}\right) \)

\( Gain(\text {Power})\;=\; 10^{\left ({\cfrac{dB(\text {Power})}{10}}\right)} \)

 

Note that, if it's not clear, where it says *Decibel values are with respect to zero reference, that means referenced to a gain of 1 which is 0dB.

 

The division by 20 is for voltage gain or current gain. For power gain you would use a factor of 10 because power is proportional to voltage or current squared.

The definition of dB is for relative power levels, where the division by 10 is used, and division by 20 is correct if you are measuring the difference in two volage or current levels at the same node.
It is also commonly used for voltage and current gain measurements from the input to output of a device, although that's only technically correct for equal input and output impedances of the device (but it's understood that requirement is mostly ignored).