# How do you make the graph for y=ln(1+x/(ln(1-x)))?

Feb 19, 2017

It looks to me like Socratic graphing cannot plot it for some reason.
This is the graph:

And my calculator can plot it also:

Feb 19, 2017

See Socratic graph and explanation.

#### Explanation:

To make y real, $1 - x > 0$, giving $x < 1$ and

$1 + \frac{x}{\ln} \left(1 - x\right) > 0$, giving

$\frac{x}{\ln} \left(1 - x\right) > - 1$, with $x < 1$.

So, if $x \in \left(0 , 1\right) , \ln \left(1 - x\right) < 0$, and so,

$x < - \ln \left(1 - x\right) \le 1$. This gives

$x \ge 1 - {e}^{- 1} = 0.63212$, nearly.

Also, the upper terminal of the graph is $\left({1}_{1} , {0}_{-}\right)$ and the lower

end is $\left(1 - \frac{1}{e} , - 1\right)$.

Now, $x \in \left[1 - \frac{1}{e} , 1\right)$, the the graph is illustrative of this aspect of the

function. I realize that the lower end is lower than $\left(1 - \frac{1}{e} , - 1\right)$.

I hope that , with the interest already shown, Steve could study my

graph{x-(e^y-1)ln(1-x)=0 [-2.5, 2.5, -1.25, 1.25]}

I've embedded my comments in the answer rather than as a comment due to all the maths:

For the domain of the function:

Condition 1

$1 - x > 0$ so that $\ln \left(1 - x\right) \in \mathbb{R} \implies \textcolor{red}{x < 1}$

Condition 2
I get a different result for the 2nd condition; as follows'

$1 + \frac{x}{\ln} \left(1 - x\right) > 0$ so that $\ln \left(1 + \frac{x}{\ln} \left(1 - x\right)\right) \in \mathbb{R}$

As we are solving an inequality we need to be careful with the sign of the denominator $\ln \left(1 - x\right)$ and change the inequality direction as appropriate.

Using the fact that for any $A \in \mathbb{R}$:

$\ln A \setminus \setminus \setminus \left\{\begin{matrix}> 0 \\ = 0 \\ < 0\end{matrix}\right. \setminus \setminus \iff \left.\begin{matrix}A > 1 \\ A = 1 \\ A < 1\end{matrix}\right.$

Lets examine the three cases:

Case 1: $\setminus \setminus \textcolor{b l u e}{\ln \left(1 - x\right) > 0}$

$\ln \left(1 - x\right) > 0 \implies 1 - x > 1 \implies x < 0$
With this requirement, we have:
$1 + \frac{x}{\ln} \left(1 - x\right) > 0$
$\therefore \ln \left(1 - x\right) + x > 0$ (as $\ln \left(1 - x\right) > 0$)
And this has no real solutions:

Case 2: $\setminus \setminus \textcolor{b l u e}{\ln \left(1 - x\right) = 0}$

$\ln \left(1 - x\right) = 0 \implies 1 - x = 1 \implies x = 0$
This case would lead to $\frac{x}{\ln} \left(1 - x\right)$ being of an indeterminate form $\frac{0}{0}$, so lets use L'Hôpital's rule to find the limit:
${\lim}_{x \rightarrow 0} \left(1 + \frac{x}{\ln} \left(1 - x\right)\right) = 1 + {\lim}_{x \rightarrow 0} \left(\frac{x}{\ln} \left(1 - x\right)\right)$
$\text{ } = 1 + {\lim}_{x \rightarrow 0} \left(\frac{1}{\frac{1}{1 - x} \left(- 1\right)}\right)$
$\text{ } = 1 - 1$
$\text{ } = 0$
and as we require $1 + \frac{x}{\ln} \left(1 - x\right) > 0$ we can eliminate this.

Case 3: $\setminus \setminus \textcolor{b l u e}{\ln \left(1 - x\right) < 0}$

$\ln \left(1 - x\right) < 0 \implies 1 - x < 1 \implies x > 0$
With this requirement, we have:
$1 + \frac{x}{\ln} \left(1 - x\right) > 0$
$\therefore \ln \left(1 - x\right) + x < 0$ (as $\ln \left(1 - x\right) < 0$)
Using WFA and graphs this has the solution $\textcolor{red}{x > 0}$

Combining all the results gives the domain of $y$ as $x \in \left(0 , 1\right)$, which is inconsistent with your analysis.

Thanks to some help from @jimh we have

${\lim}_{x \rightarrow 0} \ln \left(1 + \frac{x}{\ln} \left(1 - x\right)\right) \rightarrow - \infty$

But it does so extremely slowly due to the effect of the $\log$ function, and so the range is $y \in \left(- \infty , 0\right)$

Due to the fact that very rapidly we can be dealing with very small numbers with many decimal places, the computer accuracy is a issue when displaying the graph of the function.

Eg $A \approx 1 \times {10}^{-} 10 \implies \ln \left(A\right) \approx - 23$ so the graph appear to "tail off" in the region $- 25 < y < - 15$ but this is an illusion and for a sufficiently small $x$then $y$ decreases without bound.

Other than looking at derivatives I'm not sure there is much more to add