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

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?

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


a > 1 && x > 1

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*)

• cond is not equivalent to ieq since we don't know the sign of Log[a]. Commented Jul 13, 2023 at 7:16
• Mathematica covers both cases. Commented Jul 13, 2023 at 23:58