Answer
$$y=\sin x+1$$
Work Step by Step
To solve natural logarithm equations, keep in mind this property:
- If $\ln x = \ln a$ then $x=a$
$$\ln (y^2-1)-\ln(y+1)=\ln \sin x$$
- Condition for $y$: $y^2\gt 1$ and $y\gt-1$. That means $y\gt1$.
- Condition for $x$: $\sin x\gt0$
- Apply Quotient Rule to the left side:
$$\ln\frac{y^2-1}{y+1}=\ln \sin x$$ $$\ln\frac{(y-1)(y+1)}{y+1}=\ln \sin x$$ $$\ln (y-1)=\ln\sin x$$ $$y-1=\sin x$$ $$y=\sin x+1$$
*Reconsider the condtion:
$y\gt 1$ means $\sin x+1\gt1$ or $\sin x\gt0$
But we already limit that $\sin x\gt0$ in the condtion for $x$ above.
Thus the condition is already satisfied in the defined range of $x$.
