Question

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

Answer 1

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

Explanation:

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`