What is ln((e^y-1)/(e^y+1)) ?

The solution to ln((e^y-1)/(e^y+1)) is ln(e^y-1)-ln(e^y+1)

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simplify ln((e^y-1)/(e^y+1))

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

ln ((x - 1)^{3} (x^{2} + 1)^{4})

- ln (3) + \frac{1}{3} + 2 ln (2)

10 lo g_{e} (u) - 2 lo g_{e} (v)

- 7 ln (2) + \frac{3}{4} + 3 ln (5)

ln (x^{7} y^{9} z)

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