What is ln(e^{(nyB)/(kT)}(1+e^{-(2yB)/(kT)})^n) ?

The solution to ln(e^{(nyB)/(kT)}(1+e^{-(2yB)/(kT)})^n) is (Bny)/(kT)+nln(1+e^{-(2By)/(kT)})

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simplify ln(e^{(nyB)/(kT)}(1+e^{-(2yB)/(kT)})^n)

simplify

ln (e^{\frac{n y B}{k T}} (1 + e^{- \frac{2 y B}{k T}})^{n})

\frac{B n y}{k T} + n ln (1 + e^{- \frac{2 B y}{k T}})

Solution steps

ln (e^{\frac{n y B}{k T}} (1 + e^{- \frac{2 y B}{k T}})^{n})

Apply log rule:

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

ln (e^{\frac{n y B}{k T}} (1 + e^{- \frac{2 y B}{k T}})^{n}) = ln (e^{\frac{n y B}{k T}}) + ln ((1 + e^{- \frac{2 y B}{k T}})^{n})

= ln (e^{\frac{B n y}{k T}}) + ln ((e^{- \frac{2 B y}{k T}} + 1)^{n})

Apply log rule:

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

ln (e^{\frac{n y B}{k T}}) = \frac{n y B}{k T} ln (e)

= \frac{n y B}{k T} ln (e) + ln ((1 + e^{- \frac{2 y B}{k T}})^{n})

Apply log rule:

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

ln ((1 + e^{- \frac{2 y B}{k T}})^{n}) = n ln (1 + e^{- \frac{2 y B}{k T}})

= \frac{n y B}{k T} ln (e) + n ln (1 + e^{- \frac{2 y B}{k T}})

\frac{n y B}{k T} ln (e) = \frac{B n y}{k T}

= \frac{B n y}{k T} + n ln (e^{- \frac{2 B y}{k T}} + 1)

\frac{1}{2} ln (x - 6) - \frac{1}{12} ln (x + 6) + c

\frac{ln ( 343133 ) - ln ( 242 )}{160 - 130}

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

- lo g_{10} (424 \cdot 1 0^{(- 3)})

- 4 (ln (4) - ln (8))

What is ln(e^{(nyB)/(kT)}(1+e^{-(2yB)/(kT)})^n) ?
The solution to ln(e^{(nyB)/(kT)}(1+e^{-(2yB)/(kT)})^n) is (Bny)/(kT)+nln(1+e^{-(2By)/(kT)})