## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 1 - Section 1.6 - Inverse Functions and Logarithms - Exercises - Page 49: 54

#### 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 $x$.

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