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

1 Answer
Eddie
Jun 21, 2016

#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#

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