How do you find the y ? ln(y^2-1) - ln(y+1)=ln(sinx).

1 Answer
May 22, 2018

I'm assuming you want us to solve for #y#. Use logarithm laws.

#ln((y^2 - 1)/(y +1)) = ln(sinx)#

#ln(((y + 1)(y - 1))/(y +1)) = ln(sinx)#

#ln(y -1) = ln(sinx)#

#y - 1 =sinx#

#y = sinx +1#

However, there will be restrictions on the variable. These will occur when #sinx ≤ 0#.

Hopefully this helps!