2ln(x)<-1 — step-by-step solution

1) Let's solve in

ℝ

2 \cdot \ln (x) < -1

Divide both sides by

2

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

You can 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

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

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

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

Candidates, pending the domain check:

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

For the expression

\ln (x)

to be defined, we must have:

x > 0

Domain constraints:

x > 0

▶Solution:

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

2) The numeric value of:

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

is:

x \in (0, \frac{1}{{\approx 2.7183}^{0.5}})

x \in (0, \frac{1}{\approx 1.6487})

▷

x \in (0, \approx 0.6065)