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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

ln (\frac{e ^{- \frac{bh w}{2}}}{1 - e ^{- bh w}}) = ln (e^{- \frac{bh w}{2}}) - ln (1 - e^{- bh w})

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

Apply log rule:

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

ln (e^{- \frac{bh w}{2}}) = - \frac{bh w}{2} ln (e)

= - \frac{bh w}{2} ln (e) - ln (1 - e^{- bh w})

\frac{bh w}{2} ln (e) = \frac{bh w}{2}

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

4 (lo g_{4} (a) - lo g_{4} (b))

lo g_{10} (x (x - 2))

2 lo g_{5} (3 x) - lo g_{5} (5 x)

3 ln (x) + 5 ln (3)

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