What is ln(-1/2 e^{-2y}) ?

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

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simplify ln(-1/2 e^{-2y})

simplify

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

- 2 y - ln (2)

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

ln (e^{- 2 y}) = - 2 y ln (e)

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

Apply log rule:

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

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

= ln (1) - ln (2) - 2 ln (e) y

ln (1) = 0

2 y ln (e) = 2 y

= 0 - ln (2) - 2 y

0 - ln (2) - 2 y = - ln (2) - 2 y

= - 2 y - ln (2)

lo g_{b} (6) + lo g_{b} (10)

lo g_{10} (\frac{15}{5}) + lo g_{10} (\frac{2}{5})

- (\frac{1}{7}) lo g_{2} (\frac{1}{7}) - (\frac{6}{7}) lo g_{2} (\frac{6}{7})

\frac{1}{3} lo g_{c} (x) + 3 lo g_{c} (y) - 2 lo g_{c} (x)

- lo g_{10} (5.2 \cdot 1 0^{- 8})