log base 2

It turns out that the number 1 reason people visit this blog is to calculate log base 2 of an integer. So here is log₂ of 1 through 10, to 16 digits precision:

log₂(1) = 0
log₂(2) = 1
log₂(3) = 1.584962500721156
log₂(4) = 2
log₂(5) = 2.321928094887362
log₂(6) = 2.584962500721156
log₂(7) = 2.807354922057604
log₂(8) = 3
log₂(9) = 3.169925001442312
log₂(10) = 3.321928094887362

Note that log₂(x) is defined for any x greater than zero. If you have a calculator than computes the natural logarithm (often denoted ln), then you can calculate log₂(x) = ln(x)/ln(2). The same thing works with log base 10, i.e. log₂(x) = log₁₀(x)/log₁₀(2).

But what does it mean?
log₂(x) means the power you have to raise 2 in order to get x. For example, 2² = 4, so log₂(4) is 2. Similarly, 2³ = 8, so log₂(8) = 3. It turns out that 2^1.58496 is very nearly 3, so log₂(3) is roughly 1.58496.

Some cases deserve special mention. log₂(2) = 1 because 2¹ is 2. log₂(1) = 0 because by mathematical convention 2⁰ = 1 (this holds not just for 2, but for any base). Finally, note that log₂(0) is undefined, although some software will return -Infinity (which is the limit of log₂(x) as x approaches zero).

What is it used for?

The logarithm is useful for a variety of purposes. One of the more common is when describing exponential growth or decay. For example, the time for a radioactive substance to decay to half its mass is called the half life. Similarly we can describe accelerating growth in terms of the doubling time. I previously applied this to the number of blogs tracked by Technorati.

In computing, log₂ is often used. One reason is that the number of bits needed to represent an integer n is given by rounding down log₂(n) and then adding 1. For example log₂(100) is about 6.643856. Rounding this down and then adding 1, we see that we need 7 bits to represent 100. Similarly, in order to have 100 leaves, a binary tree needs log₂(100) levels. In the game where you have to guess a number between 1 and 100 based on whether it's higher or lower than your current guess, the average number of guesses required is log₂(100) if you use a halving strategy to bracket the answer.

Two much of nothing

Although I can't provide additional help to people with logarrhythmias, I hope this note is of some assistance.