JavaScript (ES6), 18 bytes

f=n=>n&&1+f(n>>>1)

<input type=number min=0 step=1 value=8 oninput="O.value=f(this.value)">
<input id=O value=4 disabled>

x86 Assembly, 4 bytes

Assuming Constant in EBX:

bsr eax,ebx
inc eax

EAX contains the number of bits necessary for Constant.

Bytes: ☼¢├@

Hexadecimal: ['0xf', '0xbd', '0xc3', '0x40']

Mathematica, 9 bytes

BitLength

Alternatively:

Floor@Log2@#+1&
#~IntegerLength~2&

Python, 14 bytes

int.bit_length

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C, 31 bytes

f(long n){return n?1+f(n/2):0;}

... Then I thought about recursion. From obscure to obvious, and with one fourth of the length dropped off.

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C, 43 bytes

c;
#define f(v)({for(c=0;v>=1l<<c;++c);c;})

Calling f with an unsigned value (e.g. f(42u)) will "return" its bit length. Even works for 0u !

Ungolfed and explained: (backslashes omitted)

c;
#define f(v)
    ({ // GCC statement-expression

        // Increment c until (1 << c) >= v, i.e
        // that's enough bits to contain v.
        // Note the `l` to force long
        // arithmetic that won't overflow.
        for(c = 0; v >= 1l << c; ++c)
            ; // Empty for body

        c; // Return value
    })

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Perl 6, 7 bytes

*.msb+1

Try it

Explanation:

* makes it become a WhateverCode lambda, and indicates where to put the input

.msb on an Int returns the index of the most significant bit (0 based)

+1 is combined into the lambda, and adds one to the eventual result of calling .msb.

JavaScript ES6, 19 bytes

a=>32-Math.clz32(a)

Math.clz32 returns the number of leading zero bits in the 32-bit binary representation of a number. So to get the amount of bits needed, all we need to do is substract that number from 32

f=
  a=>32-Math.clz32(a)
  

pre {
    display: inline;
}

<input id=x type="number" oninput="y.innerHTML = f(x.value)" value=128><br>
<pre>Bits needed: <pre id=y>8</pre></pre>

Jolf, 2 bytes

lB

Just convert to binary and then find the length.

Labyrinth, 13 12 bytes

 ?
_:
2/#(!@

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Explanation

The program simply repeatedly divides the input by 2 until it's zero. The number of steps are kept track of by duplicating the value at each step. Once it's reduced to zero we print the stack depth (minus 1).

The program starts at the ? which reads the input. The main loop is then the 2x2 block below, going counter-clockwise:

:   Duplicate current value.
_2  Push 2.
/   Divide.

Once the value is zero after a full iteration, the linear bit at the end is executed:

#   Push stack depth.
(   Decrement.
!   Print.
@   Terminate.

C preprocessor macro (with gcc extensions), 26

#define f 32-__builtin_clz

Uses GCC's count-leading-zeros builtin.

Call this as a function, e.g. f(100).

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Ruby, 19 16 bytes

->n{"%b"%n=~/$/}

Thanks Jordan for golfing off 3 bytes

Pyth, 3 bytes

l.B

Test suite available here.

Explanation

l.BQ    Q is implicitly appended
   Q    eval(input)
 .B     convert Q to binary string
l       len(.B(Q))

bash / Unix tools, 16 bytes

dc<<<2o$1n|wc -c

Jelly, 2 bytes

BL

Converts to binary, finds length.

Excel, 17 Bytes

Takes input from cell A1

=LEN(DEC2BIN(A1))

or

=INT(LOG(A1,2)+1)

or

=INT(LOG(2*A1,2))

Julia 0.4, 14 bytes

!n=log2(2n)÷1

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C#, 63 45 31 bytes

Saved 18 bytes, thanks to Loovjo, and TuukkaX

Saved 14 bytes, thanks to Grax

 b=>1+(int)System.Math.Log(b,2);

It uses, that a decimal number n has ⌊log2(n)⌋+1 bits, which is described on this page:

Number of Bits in a Specific Decimal Integer

A positive integer n has b bits when 2^(b-1) ≤ n ≤ 2^b – 1. For example:

• 29 has 5 bits because 16 ≤ 29 ≤ 31, or 2^4 ≤ 29 ≤ 2^5 – 1
• 123 has 7 bits because 64 ≤ 123 ≤ 127, or 2^6 ≤ 123 ≤ 2^7 – 1
• 967 has 10 bits because 512 ≤ 967 ≤ 1023, or 2^9 ≤ 967 ≤ 2^10 – 1

For larger numbers, you could consult a table of powers of two to find the consecutive powers that contain your number.

To see why this works, think of the binary representations of the integers 2^4 through 2^5 – 1, for example. They are 10000 through 11111, all possible 5-bit values.

Using Logarithms

The above method can be stated another way: the number of bits is the exponent of the smallest power of two greater than your number. You can state that mathematically as:

bspec = ⌊log2(n)⌋ + 1

That formula has three parts:

• log2(n) means the logarithm in base 2 of n, which is the exponent to which 2 is raised to get n. For example, log2(123) ≈ 6.9425145. The presence of a fractional part means n is between powers of two.

• ⌊x⌋ is the floor of x, which is the integer part of x. For example, ⌊6.9425145⌋ = 6. You can think of ⌊log2(n)⌋ as the exponent of the highest power of two in the binary representation of n.

• +1 takes the exponent to the next higher power of two. You can think of this step as accounting for the 2^0th place of your binary number, which then gives you its total number of bits. For our example, that’s 6 + 1 = 7. You might be tempted to use the ceiling function — ⌈x⌉, which is the smallest integer greater than or equal to x — to compute the number of bits as such:

bspec = ⌈log2(n)⌉

However, this fails when n is a power of two.

C#, 32 bytes

n=>Convert.ToString(n,2).Length;

Converts the parameter to a binary string and returns the length of the string.

Befunge-93, 23 21 Bytes

&>2# /# :_1>+#<\#1_.@

Befunge is a 2D grid-based language (although I'm only using one line).

&                      take integer input
 >2# /# :_             until the top of the stack is zero, halve and duplicate it
          1>+#<\#1_    find the length of the stack
                   .@  output that length as an integer and terminate the program

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Haskell, 20 bytes

succ.floor.logBase 2

Composes a function that takes logarithm base 2, floors, and adds 1.

Jellyfish, 4 bytes

p#bi

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Print (p), the length (#) of the binary representation (b) of the input (i).

Retina,  44  23 bytes

Requires too much memory to run for large input values. Converts to unary, then repeatedly divides by 2, counting how many times until it hits zero. Byte count assumes ISO 8859-1 encoding.

.*
$*
+`^(1+)1?\1
$1_
.

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CJam, 5 bytes

ri2b,

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Read input (r), convert to integer (i), get binary representation (2b), get length (,).

Octave, 19 bytes

@(x)ceil(log2(x+1))

Anonymous function that adds 1, computes binary logarithm and rounds up.

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QBIC, 18 bytes

:{~2^q>a|_Xq\q=q+1

That's incredible Mike! But how does it work?

:        Read input as integer 'a'
{        Start an infinite DO-LOOP
~2^q>a   If 2 to the power of q (which is 1 by default) is greater than a
|_Xq     Then quit, printing q
\q=q+1   Else, increment q
[LOOP is closed implicitly by QBIC]

Pyke, 3 bytes

b2l

Try it here!

Perl, 21 bytes

21 bytes, (ab)use the fact that $= must be an integer

say$==1+log(<>)/log 2

25 bytes, naïve implementation

say length sprintf"%b",<>

28 23 byte version without whitespaces

$-++while$_>>=1;say++$-

1while($i//=<>)>=1<<++$_;say

Usage

$ echo 128 | perl -E '$-++while$_>>=1;say++$-'
8
$ echo 128 | perl -E 'say length sprintf"%b",<>'
8

APL, 6 bytes

1+⌊2⍟⍵

Omega is the right argument, which is replaced with the number in question.

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Julia, 16 bytes

n->endof(bin(n))

Anonymous function.