Log Base 2 Calculator

Log base 2 answers a small question with a lot of consequences: how many times do you have to double 1 to reach x? If x is 8, the answer is 3, because $2 \times 2 \times 2 = 8$. The notation is $\log_2(x)$, and it shows up everywhere computers do, because computers count in binary. This calculator runs it both ways, feed it x to get $\log_2(x)$, or feed it the log to recover x.

How log base 2 works

$\log_2(x)$ asks: what power of 2 equals x? Plug in 16 and the answer is 4, because $2^4 = 16$. Plug in 1 and you get 0, because anything raised to the zero is 1. Fractions are normal too: $\log_2(3) \approx 1.585$, since 3 sits between $2^1$ and $2^2$. Behind the scenes the calculator uses the change-of-base trick, $\log_2(x) = \frac{\ln(x)}{\ln(2)}$, which is how most software computes it.

How to use it

Type a positive number into x and read the binary logarithm off the second field. Reverse the process by entering a value into $\log_2(x)$; the calculator returns 2 raised to that exponent. So x = 64 gives $\log_2(x) = 6$, and $\log_2(x) = 6$ gives x = 64. Either direction works.

Where binary logarithms actually show up

Algorithm analysis. When a textbook says an algorithm runs in O(log n), the log is almost always base 2. Binary search halves the problem each step, so finding an entry in a sorted list of a million items takes about 20 comparisons, $\log_2(1,000,000) \approx 19.93$. The same logic governs balanced binary trees, heaps, and merge sort.

Bit counting. To store a non-negative integer n you need $\lceil \log_2(n + 1) \rceil$ bits. That's why 255 fits in 8 bits ($\log_2(256) = 8$) and why a 32-bit unsigned integer tops out around 4.3 billion ($2^{32} = 4,294,967,296$).

Information theory. Shannon defined information in bits, and bits come straight out of log base 2. A fair coin flip carries $\log_2(2) = 1$ bit of information; a fair six-sided die carries $\log_2(6) \approx 2.585$ bits per roll. Lossless compression schemes use the same math to figure out how close they can get to a file's theoretical minimum size.

Graphics, audio, and signal processing. Mipmap levels in texture mapping step down by a factor of two per level, octaves in music double in frequency, and FFTs are happiest when input sizes are powers of two. $\log_2$ is the unit of measurement for any of those ladders.

Quick sanity checks

Powers of 2 land on whole numbers: 2, 4, 8, 16, 32, 64 give 1, 2, 3, 4, 5, 6. Anything in between is a fraction, and anything between 0 and 1 turns negative, $\log_2(0.5) = -1, \quad \log_2(0.25) = -2$ . For a rough mental estimate, count how many times you can halve the number before crossing 1. Halving 100 goes 50, 25, 12.5, 6.25, 3.125, 1.5625, 0.78… about 6.6 steps, and $\log_2(100)$ is in fact 6.644.

Frequently asked questions

How is log base 2 different from ln or log?

Different base, same idea. $\log_2$ uses 2, ln uses Euler's number $e \approx 2.718$, and bare "log" usually means base 10, except in math papers it can mean ln, and in computer science it almost always means log₂. Context decides. They are all related by constants: $\log_2(x) = \frac{\ln(x)}{\ln(2)} \approx \ln(x) \times 1.4427$.

How many bits do I need to represent a number?

For a non-negative integer n, $\lceil \log_2(n + 1) \rceil$ bits. So 0 through 255 fits in 8 bits, 0 through 65,535 fits in 16, and the pattern continues. Signed integers give up one bit for the sign, so an 8-bit signed range is -128 to 127.

What does a negative result mean?

Your input is between 0 and 1. $\log_2(0.5) = -1$ because halving 1 once gets you to 0.5; $\log_2(0.125) = -3$ because you halve three times. The logarithm is just counting doublings, and going below 1 means counting in reverse.

Why does the calculator reject zero and negative numbers?

Real logarithms only exist for positive inputs. $\log_2(0)$ heads off to negative infinity (you would need to halve forever to reach 0), and $\log_2$ of a negative number only makes sense in the complex plane, which is outside this calculator's scope.