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

Jun 21, 2016

The domain is $\left(0 , 1\right)$

#### Explanation:

The domain of the Real valued function of Reals $\ln \left(x\right)$ is $\left(0 , \infty\right)$

So looking at the denominator $\ln \left(1 - x\right)$ first, we require:

$1 - x > 0$.

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

Additionally we require $\ln \left(1 - x\right) \ne 0$ in order that the denominator be non-zero. Hence $x \ne 0$.

So $x \in \left(- \infty , 0\right) \cup \left(0 , 1\right)$

In order that $1 + \frac{x}{\ln \left(1 - x\right)} > 0$, we require:

$\frac{x}{\ln \left(1 - x\right)} > - 1$

Split into cases:

Case $x \in \left(0 , 1\right)$

Then $1 - x \in \left(0 , 1\right)$ and $\ln \left(1 - x\right) \in \left(- \infty , 0\right)$

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

$x < - \ln \left(1 - x\right)$

This inequality holds in the whole interval $\left(0 , 1\right)$. 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 $x \in \left(- \infty , 0\right)$

In this case:

$1 - x > 1$

So:

$\ln \left(1 - x\right) > 0$

Multiply both sides of the required inequality by $\ln \left(1 - x\right)$ to get:

$x > - \ln \left(1 - x\right)$

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

Proof

What is the slope of $- \ln \left(1 - x\right)$?

$\frac{d}{\mathrm{dx}} - \ln \left(1 - x\right) = \frac{1}{1 - x}$

So when $x \in \left(0 , 1\right)$ we find:

$\frac{1}{1 - x} > 1$

and when $x \in \left(- \infty , 0\right)$ we find:

$\frac{1}{1 - x} < 1$

When $x = 0$, $- \ln \left(1 - x\right) = - \ln \left(1\right) = 0$

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

Jun 21, 2016

${\log}_{e} \left(1 + \frac{x}{{\log}_{e} \left(1 - x\right)}\right)$

is feasible only for $0 < x < 1$

#### Explanation:

${\log}_{e} \left(1 + \frac{x}{{\log}_{e} \left(1 - x\right)}\right) = {\log}_{e} \left(\frac{{\log}_{e} \left(1 - x\right) + x}{{\log}_{e} \left(1 - x\right)}\right)$

The feasibility condition is

$\frac{{\log}_{e} \left(1 - x\right) + x}{{\log}_{e} \left(1 - x\right)} > 0$

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

${\log}_{e} \left(1 - x\right) + x < 0 \to {\log}_{e} \left(\frac{1}{{e}^{x} \left(1 - x\right)}\right) > 0$

or $\frac{1}{{e}^{x} \left(1 - x\right)} > 1 \to {e}^{- x} > 1 - x$

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

$1 - x \le {e}^{- x} \forall x \in \mathbb{R}$

2)Considering now the condition $1 < x < \infty$, the feasibility condition is

${\log}_{e} \left(1 - x\right) + x > 0 \to {\log}_{e} \left({e}^{x} \left(1 - x\right)\right) > 0$

or

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

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

$1 - x \le {e}^{- x} \forall x \in \mathbb{R}$

3) For $- \infty > x > 0$ we have

$1 + \frac{x}{\log} _ e \left(1 + x\right) < 0$ because $\frac{\delta}{\log} _ e \left(1 + \delta\right) \ge 1$

Concluding,

${\log}_{e} \left(1 + \frac{x}{{\log}_{e} \left(1 - x\right)}\right)$

is feasible only for $0 < x < 1$

Jun 22, 2016

$\left(0 , 1\right)$.

#### Explanation:

$\ln \left(1 - x\right)$ exists, for x in [-1, 1)#.

$\ln \left(1 + \frac{x}{\ln \left(1 - x\right)}\right)$ exists, for $\frac{x}{\ln \left(1 - x\right)} \in \left(- 1 , 1\right]$.

Now, as $x \to 0$, the given function $\to - \infty$.

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

For $x < 0 , \ln \left(1 - x\right) > 0 \mathmr{and} \frac{x}{\ln} \left(1 - x\right) < - 1$.

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

function does not exist.

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