Answer
$$y=2xe^x+1$$
Work Step by Step
To solve natural logarithm equations, keep in mind this property:
- If $\ln x = \ln a$ then $x=a$
$$\ln (y-1)-\ln2=x+\ln x$$
- 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.