(ln(1+x))^2+2ln(1+x)>0 — step-by-step solution

1) Let's solve in

ℝ

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

You can write the expression:

{\ln (x + 1)}^{2} + 2 \cdot \ln (x + 1)

in factored form:

\ln (x + 1) \cdot (\ln (x + 1) + 2)

We study the sign of each factor of the product

⇥1.1) Let's study the sign of

\ln (x + 1)

by widening the sign to ≥ and solving in

ℝ

\ln (x + 1) \geq 0

⇥Let's apply the exp function on both sides:

⇥

e^{\ln (x + 1)} \geq e^{0}

⇥ⓘ For real

x

\ln (\exp (x)) = x

; wherever

\ln (x)

is defined,

\exp (\ln (x)) = x

e^{\ln (x + 1)}

x + 1

■Any nonzero number raised to the power 0 is 1:

e^{0}

1

⇥

x + 1 \geq 1

⇥ⓘ We subtract

1

from both sides

⇥

x \geq 1 - 1

⇥ⓘ Start from

1

and move back

1

on the number line, down to zero:

1 - 1

0

⇥Candidates, pending the domain check:

x \geq 0

⇥So the expression

\ln (x + 1)

is positive or zero on

[0, +∞)

and negative otherwise

⇥1.2) Let's study the sign of

\ln (x + 1) + 2

by widening the sign to ≥ and solving in

ℝ

\ln (x + 1) + 2 \geq 0

⇥You can apply the exp function on both sides:

⇥

e^{\ln (x + 1)} \geq e^{-2}

⇥ⓘ For real

x

\ln (\exp (x)) = x

; wherever

\ln (x)

is defined,

\exp (\ln (x)) = x

e^{\ln (x + 1)}

x + 1

■The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base:

e^{-2}

\frac{1}{e^{2}}

⇥

x + 1 \geq \frac{1}{e^{2}}

⇥Candidates, pending the domain check:

x \geq \frac{1}{e^{2}} - 1

⇥So the expression

\ln (x + 1) + 2

is positive or zero on

[\frac{1}{e^{2}} - 1, +∞)

and negative otherwise

2) Let's draw up the sign table:

x	$-∞$		$\frac{1}{e^{2}} - 1$		$0$		$+∞$
$\ln (x + 1)$		-		-	0	+
$\ln (x + 1) + 2$		-	0	+		+
${\ln (1 + x)}^{2} + 2 \cdot \ln (1 + x)$		+	0	-	0	+

Candidates, pending the domain check:

x \in (-∞, \frac{1}{e^{2}} - 1) \cup (0, +∞)

3) Let's evaluate the domain of

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

For the expression

\ln (1 + x)

to be defined, we must have:

1 + x > 0

ⓘ We subtract

1

from both sides

Domain constraints:

x > -1

▶Solution:

x \in (-1, \frac{1}{e^{2}} - 1) \cup (0, +∞)

4) The numeric value of:

x \in (-1, \frac{1}{e^{2}} - 1) \cup (0, +∞)

is:

x \in (-1, \frac{1}{{\approx 2.7183}^{2}} - 1) \cup (0, +∞)

x \in (-1, \frac{1}{\approx 7.3891} - 1) \cup (0, +∞)

x \in (-1, \frac{1 - 1 \cdot \approx 7.3891}{\approx 7.3891}) \cup (0, +∞)

x \in (-1, \frac{1 - \approx 7.3891}{\approx 7.3891}) \cup (0, +∞)

x \in (-1, - \frac{\approx 6.3891}{\approx 7.3891}) \cup (0, +∞)

▷

x \in (-1, \approx -0.8647) \cup (0, +∞)