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

May 22, 2018

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

$\ln \left(\frac{{y}^{2} - 1}{y + 1}\right) = \ln \left(\sin x\right)$

$\ln \left(\frac{\left(y + 1\right) \left(y - 1\right)}{y + 1}\right) = \ln \left(\sin x\right)$

$\ln \left(y - 1\right) = \ln \left(\sin x\right)$

$y - 1 = \sin x$

$y = \sin x + 1$

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

Hopefully this helps!