ln((x-1)^2)<2 — step-by-step solution

1) Let's solve in

ℝ

\ln ({(x - 1)}^{2}) < 2

On the logarithm's domain, ln is strictly increasing:

\ln ({(x - 1)}^{2}) < 2

is equivalent to

{(x - 1)}^{2} < e^{2}

{(x - 1)}^{2} < e^{2}

An even power only depends on the absolute value:

{(x - 1)}^{2} < e^{2}

is equivalent to

| x - 1 | < e

| x - 1 | < e

| A | < B

is equivalent to

- B < A < B

- e < x - 1 < e

We add

1

to all three parts of the inequality

- e + 1 < x < e + 1

Candidates, pending the domain check:

x \in (1 - e, e + 1)

2) Let's evaluate the domain of

\ln ({(x - 1)}^{2}) < 2

For the expression

\ln ({(x - 1)}^{2})

to exist, you must check:

{(x - 1)}^{2} > 0

In ℝ, an even exponent always yields a value ≥ 0 and it is 0 only if the base is 0

3) Let's solve in

ℝ

x - 1 = 0

ⓘ We add

1

to both sides

Solution found:

x = 1

Domain constraints:

x \neq 1

▶Solution:

x \in (1 - e, 1) \cup (1, e + 1)

4) The numeric value of:

x \in (1 - e, 1) \cup (1, e + 1)

equals:

x \in (1 - \approx 2.7183, 1) \cup (1, \approx 2.7183 + 1)

▷

x \in (\approx -1.7183, 1) \cup (1, \approx 3.7183)