how to solve this inequality...??

u can remove the modulus signs here since by property both of them r always > 0 to be defined...

so it comes out:: log_xx-1 >0..

Now x is not equal to 0 and 1.
now x-1 >x⁰(=1)
so x>2

what bout x = 0,1,-1 ???????????

secondly... how can u directly remove mod...!!

please post a logical soln..!!

its logical as far as i think. ... Hav u read the properties of logarithm? If i m conceptually wrong, pleas correct me....

consider the two cases - |x| >1 and |x| <1

then proceed

I'm doing one part. Please tell where i'm wrong.

Let 0 < x < 1 (since both values can't be attained)

then log_x 1-x ≥ 0 which implies 1-x ≤ x⁰ = 1 implies x ≥ 0. but 0 < x < 1 , hence it is one solution.

Now, Let x >1

then we get x ≥2 , the other range. and then similarly for x <0

sorry @Rahul for my comment.. I have posted an incomplete solution above...

not an issue...
actually this is how the soln. goes like...!!

log _|x| |x - 1| > 0

clearly, |x| as in base > 0 and not equal to 1
so here goes the cases,

case 1) |x| > 1 so, |x - 1| > 1 (= sign is used as for 2 its possible)

now, taking the intersection we get, x E (-∞ , -1) U [2, ∞) ----------- (i)

case 2) 0 < |x| < 1 (not equal to 1 u knw why?...) so |x - 1| < 1 (don't think of x being -1)

so, taking the intersection again we get, x E (0,1)

thus, final soln. is...

x E (-∞ , -1) U (0,1) U [2, ∞) ... as far as i think..bt in AOPS ,2 is excluded but on checking it is perect