How to simultaneously determine the range of x and base a when an inequality with logarithms holds?

Question

Log[a, x + 1] > Log[a, x - 1]

the inequality above is equation holds good under all circumstances

You can see the range of x

In[52]:= ForAll[x, x > 1, Log[a, x + 1] > Log[a, x - 1]]
Resolve[%, Reals]
Reduce[%, a, Reals]

Out[52]= \!\(
\*SubscriptBox[\(\[ForAll]\), \(x, x > 1\)]\(
\*FractionBox[\(Log[1 + x]\), \(Log[a]\)] > 
\*FractionBox[\(Log[\(-1\) + x]\), \(Log[a]\)]\)\)

Out[53]= Log[a] > 0

Out[54]= a > 1

How to simultaneously determine the range of x and base a when an inequality with logarithms holds?

Answer 1

Reduce[Log[a, x + 1] > Log[a, x - 1], {a, x}, Reals]

a > 1 && x > 1

Answer 2

I don't know about simultaneous, but you can do this.

ieq = Log[a, x + 1] > Log[a, x - 1]

Looking at the result you can see Log[a] is in the denominator of each side, so multiply each side by that.

cond = MultiplySides[ieq, Log[a]]

Then Reduce the resulting set of conditions.

Reduce [cond, x]
*(Log[a] > 0 && x > 1*)