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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

lo g_{a} (1) = 0

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

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

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

10 lo g_{10} (\frac{50 \cdot 1 0 ^{- 3}}{1 \cdot 1 0 ^{- 6}})

4 ln (x) + ln (y)

ln (x) + ln (x^{2}) - 6

- \frac{1}{6} ln (12) + \frac{1}{6} ln (6)

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

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