Very Close... Your answer of 5 watts in = 970 out should be 1000 watts out. You can do this in your head.

Let's take the number ' 23 ' and break it down as 10 + 10 + 3.  The first multiplier is 10. Second multiplier is 10 again. 10 x 10 = 100. Third multiplier is 2. 100 x 2 = 200, so 5 watts into the amplifier yields 1000 watts out. If you work it backwards, 2 x 10 = 20 x 10 = 200.

Same difference.

Run 23 dB through the calculator here and you'll see the multiplier comes up as 200 for power but not voltage gain. (14.1) I don't have a shortcut for ballparking voltage gain, but where the 3 / 10 method helps is on multiple choice tests for questions like the one you asked. Almost invariably the numbers given will be multiples of 3 or 10 which makes it easy to double check the properly calculated result...

You can ballpark Decibels by remembering two values... 3 is either double or half (depending on whether it's gain or loss) and 10 dB is 10 times (or 1/10th) the original value.

In the example given, the 12 dB of gain in the antenna is reduced 6 dB by the loss in the feed line. Overall antenna system gain is now 6 dB which is 3dB + 3dB or double, then double again, for a multiplier of 4. If we want 500 watts ERP we'd need 125 watts from the transmitter. The first 3 dB of gain takes the ERP to 250 watts, the second 3 dB of gain takes the ERP to 500 watts. Or, working the reverse, 500 divided by 4 = 125.

Here's an example for you to work:  An amplifier is rated at 23 dB of gain. What's the ratio between the input and output level?

If x = b^y, then y is the logarithm of x to base b, and is written y = log_b(x). Usually the base b is 10.
In your case 500/x=10^6/10 and x= 500/10^6/10