I have come across a question while studing for my exams prove log_2x<x when x>0 I know I have to solve it using a base case eg when x=1 then assume a inductive step x=k is true but I'm having difficulty trying to solve when x=K+1 Can anyone help?

anudoneddbv

anudoneddbv

Answered question

2022-07-16

Proof by induction using logarithms
I have come across a question while studing for my exams prove
log 2 x < x  when  x > 0
I know I have to solve it using a base case eg when x = 1 then assume a inductive step x = k is true but I'm having difficulty trying to solve when x = K + 1 Can anyone help?

Answer & Explanation

Mira Spears

Mira Spears

Beginner2022-07-17Added 14 answers

Hint. Show that log ( k + 1 ) log ( k ) < ( k + 1 ) k
Matias Aguirre

Matias Aguirre

Beginner2022-07-18Added 3 answers

log 2 ( k + 1 ) < log 2 ( 2 k ) = log 2 2 + log 2 k = 1 + log 2 k < 1 + k .
The first strict inequality holds whenever k + 1 < 2 k, and that happens whenever 1 < k. So prove the result when k = 1 or k = 2 and go on from there.
You should not use both the lower-case k and the capital K interchangeably in mathematical notation. Those should be treated as if they were two separate letters. That is standard usage.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?