Q:

quick conversion of dB to $\log_2$

I was listening to this lecture
https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-451-principles-of-digital-communication-ii-spring-2005/video-lectures/lecture-1-introduction-sampling-theorem
and the prof computed $\log_2(37dB)$ in no time. I tried to follow his logic but couldn't get it. Can someone please explain how can one do such a computation in head. He used $\log_{10}(2)=0.3$ as a building block for his computation

A:

The simple logarithmic rule says that
$$\log_b(x) = \frac{\log_a(x)}{\log_a(b)}$$
so in your case,
$$\log_2(37\text{ dB}) = \frac{\log_{10} (37\text{ dB})}{\log_{10}(2)}$$
Since 37 dB = $10^{3.7}$
$$\log_2(37\text{ dB}) = \frac{\log_{10} (10^{3.7})}{\log_{10}(2)}= \frac{3.7}{\log_{10}(2)}\approx \frac{3.7}{0.3}= \frac{37}3\approx 12.3$$

