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

$$y=\frac{c-e^t}{2}$$
To solve natural logarithm equations, keep in mind this property: - If $\ln x = \ln a$ then $x=a$ $$\ln (c-2y)=t$$ - Condition: $c-2y\gt0$ or $y\lt\frac{c}{2}$ - Recall the property: $\ln e^x=x$ That means $t=\ln e^{t}$ Therefore, $$\ln(c-2y) = \ln e^{t}$$ $$c-2y=e^{t}$$ $$2y=c-e^{t}$$ $$y=\frac{c-e^t}{2}$$ We examine again the condtion that $y\lt \frac{c}{2}$: $$y=\frac{c-e^t}{2}=\frac{c}{2}-\frac{e^t}{2}$$ We know that $e^t\gt0$, so $\frac{e^t}{2}\gt0$ That means $\frac{c}{2}-\frac{e^t}{2}\lt\frac{c}{2}$ So $y\lt\frac{c}{2}$ for all $c$ already.