## Precalculus (6th Edition) Blitzer

$2012$
Let $f(x)=7$; we have $\frac{12.57}{1+4.11e^{-0.026x}}=7$ and $4.11e^{-0.026x}=\frac{12.57}{7}-1\approx0.7957$ Thus $t=-\frac{ln(0.7957/4.11)}{0.026}\approx63$, corresponding to year $1949+63=2012$