What is (ln(1+x))^2+2ln(1+x)>0 ?

The solution to (ln(1+x))^2+2ln(1+x)>0 is -1 0

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(ln(1+x))^2+2ln(1+x)>0

(ln (1 + x))^{2} + 2 ln (1 + x) > 0

- 1 < x < \frac{1}{e ^{2}} - 1 or x > 0

Interval Notation

(- 1, \frac{1}{e ^{2}} - 1) \cup (0, \infty)

Decimal Notation

- 1 < x < - 0.86466 \dots or x > 0

Solution steps

(ln (1 + x))^{2} + 2 ln (1 + x) > 0

Simplify

ln^{2} (1 + x) + 2 ln (1 + x) > 0

Let

u = ln (1 + x)

u^{2} + 2 u > 0

u^{2} + 2 u > 0 : u < - 2 or u > 0

u < - 2 or u > 0

Substitute back

u = ln (1 + x)

ln (1 + x) < - 2 or ln (1 + x) > 0

ln (1 + x) < - 2 : - 1 < x < \frac{1}{e ^{2}} - 1

ln (1 + x) > 0 : x > 0

Combine the intervals

- 1 < x < \frac{1}{e ^{2}} - 1 or x > 0

4 y + 4 = 0

x^{4} + 2 - 3

8 b = 56

∣ 2 x + 1 ∣ < 13

- 6 (v + 1) = 2 (1 - 3 v) - 7

What is (ln(1+x))^2+2ln(1+x)>0 ?
The solution to (ln(1+x))^2+2ln(1+x)>0 is -1<x< 1/(e^2)-1\lor x>0