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

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

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

simplify

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

- x - 2 ln (2)

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

ln (e^{- x}) = - x ln (e)

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

Apply log rule:

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

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

= ln (1) - ln (4) - ln (e) x

ln (1) = 0

x ln (e) = x

= 0 - ln (4) - x

0 - ln (4) - x = - ln (4) - x

= - x - ln (4)

ln (4) = 2 ln (2)

= - x - 2 ln (2)

ln (\frac{12 !}{7 ! \cdot 5 !}) - ln (11)

lo g_{10} (\frac{136.93}{10}) - 110

ln (\frac{250000000}{160141})

2 lo g_{3} (z) + 3 (lo g_{3} (w) - 2 lo g_{3} (y))

(0.0256) \cdot ln (\frac{( 5 \cdot 1 0 ^{15} )}{1.5 \cdot 1 0 ^{10}})