Log Base 2 — Binary Logarithm, Values & Examples

Log Base 2 — Binary Logarithm, Values & Examples

TL;DR

Log base 2 of a number is the power to which 2 must be raised to produce that number — so ( \log_2 8 = 3 ) because ( 2^3 = 8 ). This article covers what log base 2 (the binary logarithm) means, how to compute it by hand and with the change-of-base formula, a value table, its properties, where it powers computer science, and the slips to avoid.

What Is Log Base 2?

Log base 2 of a number ( x ), written ( \log_2 x ), is the exponent you put on 2 to get ( x ). In symbols:

[ \log_2 x = y \Longleftrightarrow 2^y = x. ]

So ( \log_2 8 = 3 ) because ( 2^3 = 8 ), and ( \log_2 32 = 5 ) because ( 2^5 = 32 ). The logarithm and the exponent are the same fact looked at from opposite ends — one asks "what is ( 2^3 )?", the other asks "2 to what gives 8?". That two-way relationship is the whole of log to exponential form conversion.

Log base 2 is also called the binary logarithm, sometimes written ( ext{lb} ) or ( ext{lg} ). The "binary" name is a hint about where it earns its keep — anything built on two states, on/off, 0/1.

How Do You Calculate Log Base 2?

There are two cases, and which one you are in decides the method.

When ( x ) is a power of 2, read the answer straight off. Ask how many times you multiply 2 to reach ( x ):

[ \log_2 16 = 4 \quad \text{since} \quad 2^4 = 16. ]

No calculator needed. The powers of 2 — 1, 2, 4, 8, 16, 32, 64, 128, 256 — are worth knowing on sight.

When ( x ) is not a power of 2, use the change-of-base formula. It states:

[ \log_2 x = \frac{\ln x}{\ln 2} = \frac{\log x}{\log 2}. ]

For example,

[ \log_2 10 = \frac{\ln 10}{\ln 2} \approx \frac{2.3026}{0.6931} \approx 3.32. ]

Log Base 2 Value Table

For the exact powers of 2, the binary logarithm is a clean whole number. This table is the one to memorise.

( x ) ( \log_2 x ) Because
1 0 ( 2^0 = 1 )
2 1 ( 2^1 = 2 )
4 2 ( 2^2 = 4 )
8 3 ( 2^3 = 8 )
16 4 ( 2^4 = 16 )
32 5 ( 2^5 = 32 )
64 6 ( 2^6 = 64 )
128 7 ( 2^7 = 128 )
256 8 ( 2^8 = 256 )
1024 10 ( 2^{10} = 1024 )

Two boundary facts close the table: ( \log_2 1 = 0 ) (anything to the power 0 is 1), and ( \log_2 0 ) is undefined — no power of 2 ever reaches 0.

What Are the Properties of Log Base 2?

Log base 2 obeys the same laws as every logarithm — they just carry a subscript 2. These come straight from the logarithm rules, specialised to base 2.

Examples of Log Base 2

Example 1

Find ( \log_2 64 ).

[ \log_2 64 = 6 ] because ( 2^6 = 64. )
Final answer: 6.

Example 2

Find ( \log_2 12 ) — and watch the tempting wrong move first.

Wrong attempt: ( \log_2 12 = 2 \times 3 = 6. ) Check it: ( 2^6 = 64 ), nowhere near 12. The answer cannot be 6.

Correct: ( \log_2 12 = \log_2(4 \times 3) = \log_2 4 + \log_2 3 = 2 + \log_2 3. )

( \log_2 3 \approx 1.585 ), so ( \log_2 12 \approx 3.585. )
Final answer: ( \log_2 12 \approx 3.585. )

Example 3

Find ( \log_2 \frac{1}{8} ).

( 1/8 = 2^{-3} ), so ( \log_2 \frac{1}{8} = -3. )
Final answer: -3.

Example 4

Use the change-of-base formula to find ( \log_2 50 ).

( \log_2 50 = \frac{\ln 50}{\ln 2} \approx 5.64. )
Final answer: ( \log_2 50 \approx 5.64. )

Example 5

Simplify ( \log_2 8 + \log_2 4 ) using the product rule.

[ \log_2 8 + \log_2 4 = \log_2(8 \times 4) = \log_2 32 = 5. ]
Final answer: 5.

Example 6

How many bits are needed to store a number from 0 to 999?

Storing ( N ) distinct values needs ( \lceil \log_2 N \rceil ) bits — the binary logarithm rounded up. With 1000 values:

[ \log_2 1000 = \frac{\ln 1000}{\ln 2} \approx 9.97, \qquad \lceil 9.97 \rceil = 10. ]

Final answer: 10 bits.

Why Log Base 2 Runs Through Computing

Log base 2 is the native tongue of computers, measuring counting bits and powering algorithms. It’s crucial in areas like:

Where Students Trip Up on Log Base 2

Common mistakes include:

  1. Forgetting the base and using base 10 by reflex.
  2. Taking the log of zero or a negative number.
  3. Multiplying logs instead of adding them.

The Short Version

Practice These Before Moving On

  1. Find ( \log_2 128 ).
  2. Find ( \log_2 \frac{1}{16} ).
  3. Use change of base to estimate ( \log_2 20 ).
  4. Simplify ( \log_2 16 - \log_2 2 ).
  5. How many bits store a number from 0 to 500?

Answers: 1. 7, 2. -4, 3. ≈ 4.32, 4. 3, 5. 9 bits.