21Let a natural number n and 10^k\leq n<10^{k+1}.
Definately (k+1) is the number of digits of n.
So, taking log on both sides we get,
k\leq \log n<k+1
Surely, k+1=\lfloor \log n+1\rfloor
can any1 prove the result that number of digits in xy is [ylogx+1]....where [] denotes G.I.F?
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