Using "log(2) is irrational", prove:For any integer number m, there does not exist maximum or minimum of log(2^m).

Answered: "Using "log(2) is irrational", prove:For any integer number..."

sublimnes9

2022-08-08

Using "log(2) is irrational", prove:
For any integer number m, there does not exist maximum or minimum of log(2^m).

Malcolm Good

Beginner2022-08-09Added 14 answers

Step 1
log(2) is irrational and using property of log:

\log (2^{m}) = m \cdot \log (2)

Step 2
For irrational value we cannot reach at any maximum or minimum value.
Hence

\log (2^{m})

has no maximum or minimum value.

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?