Precalculus (6th Edition) Blitzer

$\frac{ln3}{0.0575}\approx19.1\ years$
Using the formula $A=A_0e^{rt}$ with $A=3A_0, r=0.0575$, we have $e^{0.0575t}=3$; thus $t=\frac{ln3}{0.0575}\approx19.1\ years$