binary logarithm base 2
log₂(x) is how many times 2 must be multiplied by itself to reach x: log₂(1024)=10 because 2¹⁰=1024. It is the natural base for counting bits, levels of a binary tree and steps of an O(log n) algorithm.
Since it is not a whole number, the result rounds up: 10 bits are needed, because 2⁹=512 falls short and 2¹⁰=1024 covers all 1000 values.
2³=8 exactly, so a complete binary tree with 8 leaves has 3 levels of depth.
A frequency 100 times higher equals about 6.644 octaves above, since 2^6.644≈100.
log_b(x) = y means b^y = x, so a logarithm is simply an exponent, the power the base must be raised to. It is the inverse operation of exponentiation. For example, log₂(8) = 3 because 2³ = 8, and log₁₀(100) = 2 because 10² = 100.
⌈log₂(1000)⌉=10 bits. Since log₂(1000)≈9.9658 is not an integer, 9 bits (2⁹=512) fall short, and the result rounds up to the next integer, 10, whose 2¹⁰=1024 covers the 1000 values with room to spare.
Because log₂(8)=3 exactly, since 2³=8. Each level of a complete binary tree doubles the leaf count of the level before it, so the number of levels is always the log₂ of the total leaf count.
The log₂ of the ratio between two frequencies gives the number of octaves between them, because each octave doubles the frequency. A note 100 times higher in frequency sits log₂(100)≈6.644 octaves above, not because 100 is a round multiple of 2, but because 2^6.644 comes out close to 100.
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