bandwidth - Why Shannon Theorem has nothing to do with frequency?

According to Shannon theorem $$C= B \log _2(1 + S/N)$$ My question is, why there is no frequency in this formula?

Let's say we have 2 channels, both of them have same 20 MHz bandwidth,
one is from 100 MHz to 120 MHz,
another is from 1 GHz to 1.02 GHz.

Does this formula mean that these two channels have the same transmission speed?

Cause I think the one from 1 GHz to 1.02 GHz is faster, since it has a higher frequency.

Answer

Does this formula mean that these two channels have the same transmission speed?

Yes. That's exactly what you should take away from this: Channel capacity has nothing to do with center frequency; only bandwidth and SNR are relevant.

Cause I think the one from 1 GHz to 1.02 GHz is faster, since it has a higher frequency.

But that's plain wrong. It's not based on an understanding of what it means to use the channel:

To transport a specific amount of data, you will need to transmit something; the faster you change what you're transmitting, the more data you can send per second.

But the more often you change what you're transmitting, the higher the bandwidth. Whether that happens in a channel around 100 MHz or around 72 GHz makes no difference whatsoever.