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Mathematics 11 Online
OpenStudy (anonymous):

Prove that (log x/ log 2) + 1 <= 2 log x for all sufficiently large x.

OpenStudy (anonymous):

let lhs=(logx/log2)+1=k =>logx+log2=klog2 =>logx=(k-1)log2 =>2logx=(k-1)2log2>k for all k>1 i.e. for sufficiently large x =>2log2>(logx/log2)+1

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