CALCULATOR · MATH · LOGARITHMS

Log Base 2 Calculator

log₂(1024) = 10, because 2¹⁰ = 1024. This three-way solver goes every direction: log₂(x) for any positive x, 2^x for any exponent, and the inverse — given n, find x with log₂(x) = n (n = 20 → x = 1,048,576). The change-of-base formula log₂(x) = ln(x) ÷ ln(2) does the work, and the full powers-of-two table from 2⁰ to 2²⁰ sits below for quick reference.

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Log base 2, three ways

Pick a mode. log₂(x) answers “how many doublings reach x?” (x = 1024 → 10). 2^x does the doubling (x = 16 → 65,536). Solve log₂(x) = n inverts the first (n = 20 → 1,048,576). Fill in the field for your mode; the other field is ignored.

The 60-second rule

log₂ counts doublings. log₂(1024) = 10 because doubling 1 ten times reaches 1024. The inverse runs backward: log₂(x) = 20 means x = 2²⁰ = 1,048,576. The branch cases: log₂(0.5) = −1 (halving is a negative doubling), log₂(10) ≈ 3.3219 (10 is between 2³ = 8 and 2⁴ = 16), and 2¹⁶ = 65,536 — the old 16-bit limit that defined an era of computing.

Powers of two: 2⁰ to 2²⁰

The reference table every programmer and student eventually memorizes. Scan for the nearest value when a calculator isn’t handy.

n	2ⁿ	n	2ⁿ
0	1	11	2,048
1	2	12	4,096
2	4	13	8,192
3	8	14	16,384
4	16	15	32,768
5	32	16	65,536
6	64	17	131,072
7	128	18	262,144
8	256	19	524,288
9	512	20	1,048,576
10	1,024

The worked example, line by line

Setup: mode log₂(x), x = 1024.

Line 1 — change of base. log₂(1024) = ln(1024) ÷ ln(2) ≈ 6.9315 ÷ 0.6931 = 10.

Line 2 — check by doubling. 2¹⁰ = 1,024 — ten doublings of 1 reach exactly 1024. The table above confirms it.

Branch — a non-integer answer: x = 10 → log₂(10) = 3.3219. Since 2³ = 8 and 2⁴ = 16, the answer must sit between 3 and 4 — 3.3219 checks out.

Branch — a negative answer: x = 0.5 → log₂(0.5) = −1, because 2⁻¹ = 1/2. Halving is a “negative doubling.”

Branch — the reverse modes: 2¹⁶ = 65,536; log₂(x) = 10 → x = 2¹⁰ = 1,024.

Output · formula · meaning

Output	Formula	What it means
log₂(x)	ln(x) ÷ ln(2)	10 for x = 1024 — the number of doublings that reach x; the bit-count of the value.
2^x	2 raised to x	65,536 for x = 16 — doubling 1 exactly x times.
x with log₂(x) = n	2ⁿ	1,048,576 for n = 20 — how many distinct values n bits can name.

Who uses this

The programmer sizing storage: log₂(1048576) = 20 — a million items fits in a 20-bit address. Every bit doubles capacity, and this page converts between the two in one click.
The student learning log rules: checks log₂(10) = 3.3219 against the table — seeing 2³ = 8 < 10 < 16 = 2⁴ makes the non-integer answer click.
The audio engineer or musician: an octave is exactly a log₂ of 1 — doubling frequency. Two octaves up from 440 Hz is 440 × 2² = 1,760 Hz.
The data scientist or ML student: entropy is measured in bits = log₂ of the possibilities — 8 equally likely outcomes carry exactly 3 bits of information.

Accuracy & limitations

Results come from the change-of-base formula log₂(x) = ln(x) ÷ ln(2) computed in double-precision floating point, rounded to 4 decimal places. Very near a power of two, floating-point arithmetic can return 9.9999 or 10.0001 instead of exactly 10 — that’s the computer’s arithmetic, not your math. log₂ is undefined for x ≤ 0; 2^x guards at ±1024 because bigger exponents overflow to infinity and smaller ones round to zero. The powers table is exact for 2⁰ through 2²⁰. This is a reference calculator, not a CAS — symbolic work still belongs in a classroom or a computer algebra system.

Cite this calculatorBright Side Kit. “Log Base 2 Calculator: Solve log2(x) | Bright Side Kit.” brightoolbox.com, 2026.

Bright Side Kit. "Log Base 2 Calculator: Solve log2(x) | Bright Side Kit." https://brightoolbox.com/log-base-2-calculator/ Accessed October 24, 2026.

Data sources

Change-of-base identity — log₂(x) = ln(x) ÷ ln(2), the formula every calculator on this page evaluates.
Powers-of-two table (2⁰–2²⁰) — computed values; exact integers through 2²⁰ = 1,048,576.

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Frequently asked questions

What is log base 2?

The number of times you double 1 to reach x. log₂(1024) = 10 because 2 × 2 × … (ten times) = 1024. It’s the logarithm that counts in doublings instead of tens, which is why computing runs on it.

Why does 1024 = 2^10 matter so much?

It’s the kilobyte’s birthday. 2¹⁰ = 1,024 — close enough to 1,000 that computer people named 1,024 bytes a kilobyte. Twenty of them make 2²⁰ = 1,048,576 (a megabyte), and the powers table above is the same doubling ladder, rung by rung.

How do log2, ln, and log10 differ?

Only the base. log₂ counts doublings (base 2), ln counts e-foldings (base e ≈ 2.718), log₁₀ counts tens. They convert exactly: log₂(x) = ln(x) ÷ ln(2) = log₁₀(x) ÷ log₁₀(2). Same information, different yardstick.

Why is log2(0) undefined?

No power of 2 ever reaches 0. 2^x is always positive — positive, smaller, tinier, but never zero and never negative. Asking “2 to what power is 0?” has no answer, so log₂(0) and log₂ of any negative are undefined.

How many bits do I need for N values?

ceil(log₂(N)). Round the log up to the next whole number: 1,000 values need ceil(log₂(1000)) = ceil(9.97) = 10 bits. That’s exactly what the “solve log₂(x) = n” mode does in reverse — it names how many values n bits hold.

Why does my answer come out 9.9999 instead of 10?

Floating-point rounding. The change-of-base formula divides two irrational logs, so near a power of two the result can land a hair off the integer. The math is right; the display shows the raw double. Round to the nearest whole number when you’re clearly at a power of two.