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

$$y=2xe^x+1$$
To solve natural logarithm equations, keep in mind this property: - If $\ln x = \ln a$ then $x=a$ $$\ln (y-1)-\ln2=x+\ln x$$ - Condition: $y\gt1$ and $x\gt 0$ - Apply Quotient Rule to the left side: $$\ln\frac{y-1}{2}=x+\ln x$$ - For $x$, recall the property: $\ln e^x=x$ That means $$\ln\frac{y-1}{2}=\ln e^x+\ln x$$ - Apply Product Rule to the right side: $$\ln\frac{y-1}{2}=\ln xe^x$$ $$\frac{y-1}{2}=xe^x$$ $$y=2xe^x+1$$ *Reconsider the condtion: $y\gt 1$ means $2xe^x+1\gt1$ or $xe^x\gt0$ As stated above in the condition, we know that $x\gt0$. In addition, $e^x\gt0$ for all $x\in R$. Therefore, $xe^x\gt0$ for all $x$ in the defined range already.