How to perform multiplication, using bitwise operators?

To multiply by any value of 2 to the power of N (i.e. 2^N) shift the bits N times to the left.

0000 0001 = 1 

times 4 = (2^2 => N = 2) = 2 bit shift : 0000 0100 = 4

times 8 = (2^3 -> N = 3) = 3 bit shift : 0010 0000 = 32

etc..

To divide shift the bits to the right.

The bits are whole 1 or 0 - you can't shift by a part of a bit thus if the number you're multiplying by is does not factor a whole value of N ie.

since: 17 = 16  + 1 
thus:  17 = 2^4 + 1

therefore: x * 17 = (x * 16) + x in other words 17 x's  

thus to multiply by 17 you have to do a 4 bit shift to the left, and then add the original number again:

==> x * 17 = (x * 16) + x 
==> x * 17 = (x * 2^4) + x 
==> x * 17 = (x shifted to left by 4 bits) + x 

so let x = 3 = 0000 0011 

times 16 = (2^4 => N = 4) = 4 bit shift : 0011 0000 = 48

plus the x (0000 0011)

ie.

    0011 0000  (48)  
+   0000 0011   (3)
=============
    0011 0011  (51)

Edit: Update to the original answer. Charles Petzold has written a fantastic book 'Code' that will explain all of this and more in the easiest of ways. I thoroughly recommend this.

To multiply two binary encoded numbers without a multiply instruction. It would be simple to iteratively add to reach the product.

unsigned int mult(x, y)
unsigned int x, y;
{
    unsigned int reg = 0;

    while(y--)
        reg += x;
    return reg;
}

Using bit operations, the characteristic of the data encoding can be exploited. As explained previously, a bit shift is the same as multiply by two. Using this an adder can be used on the powers of two.

// multiply two numbers with bit operations

unsigned int mult(x, y)
unsigned int x, y;
{
    unsigned int reg = 0;

    while (y != 0)
    {
        if (y & 1)
        {
            reg += x;
        }
        x <<= 1;
        y >>= 1;
    }
    return reg;
}