What is ln(1/(4x)*e^{(-1)/2}) ?

The solution to ln(1/(4x)*e^{(-1)/2}) is -ln(x)-2ln(2)-1/2

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simplify ln(1/(4x)*e^{(-1)/2})

simplify

ln (\frac{1}{4 x} \cdot e^{\frac{- 1}{2}})

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

Solution steps

ln (\frac{1}{4 x} e^{\frac{- 1}{2}})

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

ln (\frac{1}{4 x} e^{\frac{- 1}{2}}) = ln (\frac{1}{4 x}) + ln (e^{\frac{- 1}{2}})

= ln (\frac{1}{4 x}) + ln (e^{\frac{- 1}{2}})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

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

= ln (\frac{1}{4 x}) + \frac{- 1}{2} ln (e)

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

ln (\frac{1}{4 x}) = ln (1) - ln (4 x)

= ln (1) - ln (4 x) + \frac{- 1}{2} ln (e)

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

ln (4 x) = ln (4) + ln (x)

= ln (1) - (ln (x) + ln (4)) + \frac{- 1}{2} ln (e)

ln (1) = 0

- (ln (4) + ln (x)) = - (2 ln (2) + ln (x))

\frac{- 1}{2} ln (e) = - \frac{1}{2}

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

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

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

- (ln (x) + 2 ln (2)) : - ln (x) - 2 ln (2)

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

4 ln (1 + 2)

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What is ln(1/(4x)*e^{(-1)/2}) ?
The solution to ln(1/(4x)*e^{(-1)/2}) is -ln(x)-2ln(2)-1/2