Prove that if 2k n < 2k+1 then :log n; + 1 =
Chapter 5, Problem 36(choose chapter or problem)
Prove that if \(2^k \leq n < 2^{k+1}\) then \(\lfloor\log n\rfloor+1=\lceil\log (n+1)\rceil\) . Here log n means \(\log _2 n .)\)
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