How do you solve $ln(y^2-1)-ln(y+1)=ln(sinx)$ ?

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Answer 1 · 2016-06-21T15:53:11

$color{red}{\mathbf{y \ne -1}}, y -1 = sin x \implies y = sin x + 1$

consolidate the LHS to get

$ln( (y^2 -1)/(y+1) ) = ln (sin x)$

$(y^2 -1)/(y+1) = sin x$

$((y -1)(y+1))/(y+1) = sin x$

$color{red}{\mathbf{y \ne -1}}, y -1 = sin x \implies y = sin x + 1$

How do you solve ln(y^2-1)-ln(y+1)=ln(sinx)ln(y^2-1)-ln(y+1)=ln(sinx)?