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

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

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

simplify

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

x + 2 y - ln (1 - e^{x + 2 y})

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

ln (e^{x + 2 y}) = (x + 2 y) ln (e)

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

(x + 2 y) ln (e) = x + 2 y

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

- lo g_{10} (\frac{1 0 ^{- 14}}{2 \cdot 1 0 ^{- 9}})

ln (\frac{4.67 \cdot 1 0 ^{22}}{5 \cdot 1 0 ^{22}})

\frac{1}{6} ln ∣ sin (6 x) ∣ - \frac{1}{6} ln ∣ cos (6 x) ∣

ln (\frac{x ^{2} x - 1}{3 x - 9})

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