Prove the inequality for all natural numbers n using induction log_2 n<n I know how to prove the base case Base Case log_2 1<1 likewise assuming the inequality for n=k; log_2 k<k Then to prove by induction I show log_2 k<(k+1)?

Nigro6f 2022-10-19 Answered
Prove the inequality for all natural numbers n using induction
log 2 n < n
I know how to prove the base case Base Case log 2 1 < 1 likewise assuming the inequality for n=k; log 2 k < k
Then to prove by induction I show log 2 k < ( k + 1 )?
I know it's true since the domain is all real numbers i just cant figure out the next step to prove it.
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Answers (2)

Liam Everett
Answered 2022-10-20 Author has 16 answers
log 2 ( n ) = ln ( n ) ln ( 2 ) , where l n is the natural log i.e. with base e
You need to show, log 2 ( n ) < n ln ( n ) ln ( 2 ) < n ln ( n ) < n ln ( 2 ) ln ( n ) < ln ( 2 n )
Since log is an increasing function you can rephrase your question to n < 2 n
That is easy to show by induction
n=1: 1 < 2 1 . True.
Induction Hypothesis: n 1 < 2 n 1
for n 2, n 2 ( n 1 ) < 2 2 n 1 = 2 n
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tikaj1x
Answered 2022-10-21 Author has 4 answers
For k > 1, we have
k + 1 < 2 k ,
hence, because log ( x ) is increasing
log 2 ( k + 1 ) < log 2 ( 2 k ) = log 2 ( k ) + 1 < k + 1
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