# How show log base 2 (3) is in (1,2)?

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### 2 Answers

You need to test if `log_2 (3)` belongs to the interval `(1,2)` , hence, you need to check if the following inequality `1 < log_2 (3) < 2` holds, such that:

`1 <log_2 (3) < 2`

Replacing `log_2 (2)` for 1 yields:

`log_2 (2) < log_2 (3) < 2log_2 (2)`

Using the logarithmic identity `n*log_a (b) = log_a (b^n)` yields:

`log_2 (2) < log_2 (3) < log_2 (2^2)`

`log_2 (2) < log_2 (3) < log_2 (4)`

Since the base of logarithm is larger than 1, hence, the logarithmic function increases and the inequality preserves its direction, such that:

`log_2 (2) < log_2 (3) < log_2 (4) => 2 < 3 < 4`

Since the last inequality represents a valid statement, yields that `1 <log_2 (3) < 2` .

**Hence, testing if the statement `log_2 (3) in (1,2)` is valid, using the property of logarithmic function, yields that the `log_2 (3) in (1,2)` holds.**

Show that `log_2 3` is in the interval (1,2):

Logarithms are everywhere (monotonically) increasing functions (on their domains) so `a<b => loga < logb` .

`log_2 2=1` since `2^1=2` and

`log_2 4=2` since `2^2=4`

Thus `log_2 2 < log_2 3 < log_2 4` or equivalently

`1<log_2 3 < 2` as required.