What is ln(y)-1/2 (ln(1+y)-ln(1-y)) ?

The solution to ln(y)-1/2 (ln(1+y)-ln(1-y)) is ln((ysqrt(1-y)sqrt(1+y))/(1+y))

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simplify ln(y)-1/2 (ln(1+y)-ln(1-y))

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

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

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

Apply log rule:

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

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

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

Simplify

\frac{y}{( \frac{1 + y}{1 - y} ) ^{\frac{1}{2}}} : \frac{y 1 - y 1 + y}{1 + y}

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

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What is ln(y)-1/2 (ln(1+y)-ln(1-y)) ?
The solution to ln(y)-1/2 (ln(1+y)-ln(1-y)) is ln((ysqrt(1-y)sqrt(1+y))/(1+y))