2ln(-x)<1 — step-by-step solution | Explain MAths

2ln(-x)<1

1) Let's solve in

ℝ

:

2 \cdot \ln (- x) < 1

Divide both sides by

2

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

Let's apply the exp function on both sides:

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

ⓘ For real

x

,

\ln (\exp (x)) = x

; wherever

\ln (x)

is defined,

\exp (\ln (x)) = x

:

e^{\ln (- x)}

=

- x

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

ⓘ We add

x

and subtract

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

from both sides

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

Candidates, pending the domain check:

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

2) Let's find the domain of

2 \cdot \ln (- x) < 1

For the expression

\ln (- x)

to exist, you must check:

- x > 0

ⓘ We add

x

to both sides

0 > x

Domain constraints:

x < 0

▶Solution:

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

3) The numeric value of:

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

equals:

x \in (- {\approx 2.7183}^{0.5}, 0)

▷

x \in (\approx -1.6487, 0)