What is ln((1/((2x)))e^{(-|a|)/x}) ?

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

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simplify ln((1/((2x)))e^{(-|a|)/x})

simplify

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

- \frac{∣ a ∣}{x} - ln (x) - ln (2)

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

ln (e^{\frac{- ∣ a ∣}{x}}) = \frac{- ∣ a ∣}{x} ln (e)

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

Apply log rule:

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

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

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

Apply log rule:

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

ln (2 x) = ln (2) + ln (x)

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

ln (1) = 0

\frac{- ∣ a ∣}{x} ln (e) = - \frac{∣ a ∣}{x}

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

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

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

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

= - \frac{∣ a ∣}{x} - ln (x) - ln (2)

2 lo g_{b} (z) + 3 (lo g_{b} (x) - 4 lo g_{b} (w))

- [\frac{3}{5} lo g_{2} (\frac{3}{5}) + \frac{2}{5} lo g_{2} (\frac{2}{5})]

ln (4.98 \cdot 1 0^{15} \cdot e^{0.47 (2000 - 2010)})

lo g_{10} (6) lo g_{10} (2) + lo g_{10} (3)

(\frac{1}{0.143}) ln (\frac{6 \cdot 1 0 ^{9}}{29000})

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