Maximum number of digits in numbers between 0 to n 2 − 1 of base nThe number of digits in numbers between 0 and n 2 − 1 of base n is obtained by log n ⁡ ( n 2 ) = 2 log n ⁡ n = 2But why log is being used? I mean how doing log gives correct answer always?

Question

Maximum number of digits in numbers between 0 to       n    2    −  1 of base nThe number of digits in numbers between 0 and       n    2    −  1 of base n is obtained by      log    n    ⁡  (      n    2    )  =  2      log    n    ⁡  n  =  2But why log is being used? I mean how doing log gives correct answer always?

Melanie Lam · Accepted Answer

In base   n,   n  is represented as       10    n  . So      10    n    2    =      100    n  That&#039;s 3 digits.Now   1  +      log    n    ⁡  (      n    2    )  =  3. Done.

Jamir Summers · Accepted Answer

About the number of digits of a number       n    2   in base   b:Let n have the representation  n  =  (      d          m      −      1        ⋯      d    1        d    0        )    b    =      ∑          k      =      0              m      −      1            d    k        b    k  with   m digits from   {  0  ,  …  ,  b  −  1  }. Then we have  n  &amp;lt;      b    m    ⇒      n    2    &amp;lt;      b          2      m      This means       n    2   has at most   2  m  −  1 digits.To derive m for a given n we use some logarithm:  n  &amp;lt;      b    m    ⇒    log  ⁡  n  &amp;lt;  log  ⁡      b    m    =  m  log  ⁡  b  ⇒            log      ⁡      n              log      ⁡      b        &amp;lt;  mThe smallest m should be  m  =      ⌊                  log        ⁡        n                    log        ⁡        b              ⌋    +  1The special case   b  =  n gives   m  =  2 and that       n    2   has at most   3 digits.As pointed out by fellow users       n    2    =  (  100      )    n   has exactly   3 digits, which means that your formula, which gives   2 digits, is not correct.

Maximum number of digits in numbers between 0 to n^2−1 of base n log_n(n^2) = 2 log_n n = 2

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