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

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

And my calculator can plot it also:

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

#1+x/ln(1-x)>0#, giving

#x/ln(1-x) > -1#, with #x < 1#.

So, if #x in (0, 1), ln(1-x)<0#, and so,

#x<-ln(1-x)<=1#. This gives

#x>=1-e^(-1)=0.63212#, nearly.

Also, the upper terminal of the graph is #(1_1, 0_-)# and the lower

end is #(1-1/e, -1)#.

Now, #x in [1-1/e, 1)#, the the graph is illustrative of this aspect of the

function. I realize that the lower end is lower than #(1-1/e, -1)#.

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

answer and give his comments.

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

@a-s-adikesavan

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(1-x) in RR => color(red)(x<1)#
This is consistent with your answer

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

#1+x/ln(1-x) > 0# so that #ln(1+x/ln(1-x)) in RR#

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

Using the fact that for any #A in RR#:

# ln A \ \ \ { (>0), (=0), (<0) :} \ \ iff {: (A>1), (A=1), (A<1) :}#

Lets examine the three cases:

Case 1: #\ \ color(blue)(ln(1-x)>0)#

#ln(1-x)>0 =>1-x>1=>x<0 #
With this requirement, we have:
#1+x/ln(1-x) > 0 #
# :. ln(1-x)+x>0 # (as #ln(1-x)>0#)
And this has no real solutions:

Case 2: #\ \ color(blue)(ln(1-x)=0)#

#ln(1-x)=0 =>1-x=1=>x=0 #
This case would lead to #x/ln(1-x)# being of an indeterminate form #0/0#, so lets use L'Hôpital's rule to find the limit:
#lim_(x rarr 0)(1+x/ln(1-x)) = 1+lim_(x rarr 0)(x/ln(1-x))#
#" "= 1+lim_(x rarr 0)(1/(1/(1-x)(-1)))#
#" "= 1-1#
#" "= 0#
and as we require #1+x/ln(1-x)>0# we can eliminate this.

Case 3: #\ \ color(blue)(ln(1-x)<0)#

#ln(1-x)<0 =>1-x<1=>x>0 #
With this requirement, we have:
#1+x/ln(1-x) > 0 #
# :. ln(1-x)+x<0 # (as #ln(1-x)<0#)
Using WFA and graphs this has the solution #color(red)(x>0)#

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

Thanks to some help from @jimh we have

# lim_(x rarr 0)ln(1+x/ln(1-x)) rarr -oo#

But it does so extremely slowly due to the effect of the #log# function, and so the range is #y in (-oo,0)#

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~~1xx10^-10 => ln(A) ~~ -23# so the graph appear to "tail off" in the region #-25 lt y lt -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