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Answer:
Step-by-step explanation:
Squaring on both sides
log₂ [(2x-3)/(x-1)] < 1
keeping these as powers of two
2^(log₂ [(2x-3)/(x-1)]) < 2^1
[(2x-3)/(x-1)] < 2
subracting with 2 on both sides and doing lcm
we get
1/(1-x) < 0
as 1 is positive number
So 1-x <0
there fore x > 0 -----> 1
But not only that in log there should be only +ve numbers and under should also have positive numbers by this we get
[(2x-3)/(x-1)] > 1
subtrating one on both sides and taking lcm we get
[(x-2)/(x-1)] > 0
it implies that x-2 and x-1 should have same sign
it satisfies for x<1 or x>2 ------> 2
From 1 and 2 we get
x belongs to (0,1) U (2,∞)
THANK YOU
HOPE IT IS HELPFUL
IF ANY DOUBT ASK IN COMMENT SECTION
MARK ME AS BRAINLIEST BECAUSE I HAVE TIRED A LOT BY SOLVING THIS