# Calculate: logarithm base(2x+1) of(3-2x)<1

### 1 Answer | Add Yours

You need to replace `log_(2x + 1) (2x + 1)` for 1 in the given inequality, such that:

`log_(2x + 1) (3 - 2x) < log_(2x + 1) (2x + 1)`

You need to consider two cases, such that:

If the base `(2x + 1)<1` , the logarithmic function decreases, such that:

`log_(2x + 1) (3 - 2x) < log_(2x + 1) (2x + 1) => 3 - 2x > 2x + 1`

`-2x - 2x > 1 - 3 => -4x > -2 => x < 1/2`

The logarithmic functions hold for the following values of x, such that:

`{(3 - 2x > 0),(2x + 1>0):}` => `{(x<3/2),(x>-1/2):} => x in (-1/2,3/2)`

The inequality holds, when the base `(2x + 1)<1` , for `x in (-oo,1/2) nn (-1/2,3/2) = (-1/2,1/2).`

If the base `(2x + 1)>1` , the logarithmic function increases, such that:

`log_(2x + 1) (3 - 2x) < log_(2x + 1) (2x + 1) => 3 - 2x < 2x + 1`

`-4x < -2 => x > 1/2`

The inequality holds, when the base ` (2x + 1)>1` , for `x in (1/2,oo) nn (-1/2,3/2) = (1/2,3/2).`

**Hence, collecting the answers yields that the inequality depends on the nature of base, hence, if `(2x + 1)>1` , then `x in (1/2,3/2)` and i`f (2x + 1)<1` , then `x in (-1/2,1/2).` **