Log_Tricks

Quick Trick:
log2(5)=2;    //O based index of MSB of 5 in Binary(Base2)
log10(5)+1=1; //0 based index of MSB if 10 in Base 10=0. Adding 1 will give us the number of digits of 5 in base 10.

Explanation:

Trick1:
logBASE(n)   returns the index(0 based) of the MSB of n

Example:

        log2(32)=5

            32 in Binary          :  1 0 0 0 0 0
            0 based reversed index   5 4 3 2 1 0

        so log2(32) returns the index of the MSB of n in given base(2 here)

Trick2: 

     Derivation: 
         Since logBASE(n) returns the index of the MSB of n in given base this can be further used to calculate number of digits in a number.How?
         Number of digits in a number is the index(1based) at which its MSB is present.

       Example:

         in base 2 
              n=32
              log2(32)=5;

              32: 1 0 0 0 0 0
         Index:   5 4 3 2 1 0 
         MSB located at (5th Index in zero based indexing) & (6th Index in 1 based indexing);
         so No. of digits will be O based indexing +1

         in base 10

           n=1000
           log10(1000)=3
         Here also MSB is located at 3rd index (0based) & 3+1=4th Index 1 based.

 As you can see the 1 based index of the MSB is equal to the number of digts of that number,so we have derived this formula:

 logBASE(n)+1       returns no. of digits of n in given BASE