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

$$y=\ln\Big(\frac{x}{3}\Big)$$
$$9e^{2y}=x^2$$ Take the natural logarithm of both sides: $$\ln(9e^{2y})=\ln(x^2)$$ Remember that $\ln x^a=a\ln x$ and $\ln(xy)=\ln x+\ln y$ $$\ln9+2y(\ln e)=2\ln x$$ $$\ln9+2y=2\ln x$$ $$2y=2\ln x-\ln 9$$ $$y=\ln x-\frac{\ln9}{2}$$ We can rewrite $\ln9=\ln3^2=2\ln3$. $$y=\ln x-\frac{2\ln3}{2}$$ $$y=\ln x-\ln3=\ln\Big(\frac{x}{3}\Big)$$