please help in this calculus question!!!?

1.
ln(x) ≥ ln(2x - 1) + 1
ln(x) ≥ ln(2x - 1) + ln(e)
ln(x) ≥ ln[e(2x - 1)]
x ≥ 2ex - e
x - 2ex ≥ -e
2ex - x ≤ e
(2e - 1)x ≤ e
x ≤ e/(2e - 1) …… {1}

As log(a negative number) is undefined.
x > 0 and 2x - 1 > 0
x > 0 and 2x > 1
x > 0 and x > (1/2)
Then x > 1/2 …… {2}

Combine {1} and {2}
x ≤ e/(2e - 1) and x > 1/2
(1/2) < x ≤ e/(2e - 1)


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ln[(1 - 2x)/(x - 1)] ≥ 0
ln[(1 - 2x)/(x - 1)] ≥ ln(1)
(1 - 2x)/(x - 1) ≥ 1
-(1 - 2x)/(x - 1) ≤ -1
[(2x -1)/(x - 1)] + 1 ≤ 0
[(2x -1)/(x - 1)] + [(x - 1)/(x - 1)] ≤ 0
[(2x - 1) + (x - 1)]/(x - 1) ≤ 0
(3x - 2)/(x - 1) ≤ 0
(3x - 2)(x - 1)/(x - 1)² ≤ 0
(3x - 2)(x - 1) ≤ 0 and x - 1 ≠ 0
2/3 ≤ x ≤ 1 and x ≠ 1
Then, 2/3 ≤ x < 1

When (1 - 2x)/(x - 1) ≥ 1, (1 - 2x)/(x - 1) must be positive.
Then, in the range of 2/3 ≤ x < 1, ln[(1 - 2x)/(x - 1)] must be defined.

Hence, the answer is: 2/3 ≤ x < 1