Question

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

Answer 1

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!