Logarithmic inequality

[surreal maple]

$\log_a(a^x+1)+ \frac{1}{\log{a^x-1}(a)} \le x-1 +\log_a(a^2-1)$
if $x\in(0,1]$, find the range of real parameter a.

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[sudden cloud]

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[surreal maple]

I have reached this part $a^{2x}-1-a^{x+1}+a^{x-1}\le 0$
I also found out $a>$1 for defining the logs and x not equal to $log_a(2)$ Answer looks like $1<a<\infty$ however for some reason it is $1<a<2$, I'm struggling to simplify further from this point

[fathom saffron]

\infty

[surreal maple]

?

[fathom saffron]

it's the latex for infinity

[sudden cloud]

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[surreal maple]

Thanks.

$a\in(1,\infty)$ is what I got

[sudden cloud]

Currently Busted

[fathom saffron]

can you elaborate on the 2nd line

[surreal maple]

I mean the answer given is 1<a<2 but I only got a>1

x≠log_a(2)