Question

How do you find the interval of existence for the real function `ln(1+x/(ln(1-x)))`?

Answer 1

The domain is `(0, 1)`

Explanation:

The domain of the Real valued function of Reals `ln(x)` is `(0, oo)`

So looking at the denominator `ln(1-x)` first, we require:

`1 - x > 0`.

Add `x` to both sides to find `1 > x`, that is `x < 1`

Additionally we require `ln(1-x) != 0` in order that the denominator be non-zero. Hence `x != 0`.

So `x in (-oo, 0) uu (0, 1)`

In order that `1+x/(ln(1-x)) > 0`, we require:

`x/(ln(1-x)) > -1`

Split into cases:

Case `x in (0, 1)`

Then `1-x in (0, 1)` and `ln(1-x) in (-oo, 0)`

Multiply both sides of the inequality by `ln(1-x)` and reverse the inequality (since `ln(1-x) < 0`) to get:

`x < -ln(1-x)`

This inequality holds in the whole interval `(0, 1)`. I may prove later, but here's a plot for now:

graph{(y+ln(1-x))(y-x)=0 [-1.895, 3.105, -0.68, 1.818]}

Case `x in (-oo, 0)`

In this case:

`1 - x > 1`

So:

`ln(1-x) > 0`

Multiply both sides of the required inequality by `ln(1-x)` to get:

`x > -ln(1-x)`

This is false for all `x < 0` so there is no `x < 0` in the domain.

Proof

What is the slope of `-ln(1-x)`?

`d/(dx) -ln(1-x) = 1/(1-x)`

So when `x in (0, 1)` we find:

`1/(1-x) > 1`

and when `x in (-oo, 0)` we find:

`1/(1-x) < 1`

When `x = 0`, `-ln(1-x) = -ln(1) = 0`

So `-ln(1-x)` is steeper than `y=x` when `x > 0` and less steep when `x < 0`. They both pass through `(0, 0)`.

Answer 2

`log_e(1+x/(log_e(1-x)))`

is feasible only for `0 < x < 1`

Explanation:

`log_e(1+x/(log_e(1-x)))=log_e((log_e(1-x)+x)/(log_e(1-x)))`

The feasibility condition is

`(log_e(1-x)+x)/(log_e(1-x)) > 0`

`1)` For `0 < x < 1` we have the condition

`log_e(1-x)+x < 0->log_e (1/(e^x(1-x))) > 0`

or `1/(e^x(1-x)) > 1->e^{-x} > 1-x`

but `y=1-x` is tangent to `y=e^{-x}` at `x = 0` and truly

`1-x <= e^{-x} forall x in RR`

`2)`Considering now the condition `1 < x < oo`, the feasibility condition is

`log_e(1-x)+x > 0->log_e(e^x(1-x)) > 0`

or

`1-x>e^{-x}`

but `y=1-x` is tangent to `y=e^{-x}` at `x = 0` and truly

`1-x <= e^{-x} forall x in RR`

`3)` For `-oo > x > 0` we have

`1+x/log_e(1+x) < 0` because `delta/log_e(1+delta) >=1`

Concluding,

`log_e(1+x/(log_e(1-x)))`

is feasible only for `0 < x < 1`

Answer 3

`( 0, 1 )`.

Explanation:

`ln(1-x)` exists, for x in [-1, 1)#.

`ln(1+x/(ln(1-x)))` exists, for `x/(ln(1-x)) in (-1, 1]`.

Now, as `x to 0`, the given function `to -oo`.

The proof is a tribute to L'Hospital rule..

For `x < 0, ln(1-x)>0 and x/ln(1-x)<-1`.

So, the argument for the given ln function becomes negative. The

function does not exist.

Thus, the answer is (0, 1).. .