log(1/2,x)>1 — step-by-step solution

1) Let's solve in

ℝ

\log (\frac{1}{2}, x) > 1

ⓘ log(b, v) is the logarithm of v in base b, that is ln(v)/ln(b):

\log (\frac{1}{2}, x)

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

\frac{\ln (x)}{\ln (\frac{1}{2})} > 1

2) Let's examine the sign of:

\ln (\frac{1}{2})

\ln (0.5)

\approx -0.6931

We multiply both sides by

\ln (\frac{1}{2})

; since this number is negative, we reverse the inequality sign

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

ⓘ We subtract

\ln (\frac{1}{2})

from both sides

\ln (x) - \ln (\frac{1}{2}) < 0

ⓘ For strictly positive arguments, a sum of logarithms is the logarithm of the product, a difference that of the quotient, ln(a) + ln(b) = ln(a*b):

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

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

\ln (\frac{x}{\frac{1}{2}}) < 0

ⓘ Dividing is multiplying by the reciprocal:

\frac{x}{\frac{1}{2}}

x \cdot 2

\ln (x \cdot 2) < 0

You can apply the exp function on both sides:

e^{\ln (x \cdot 2)} < e^{0}

ⓘ For real

x

\ln (\exp (x)) = x

; wherever

\ln (x)

is defined,

\exp (\ln (x)) = x

e^{\ln (x \cdot 2)}

x \cdot 2

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

e^{0}

1

x \cdot 2 < 1

ⓘ Both sides are divided by

2

Candidates, pending the domain check:

x < \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}{2})

3) The numeric value of:

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

is:

▷

x \in (0, 0.5)