$\dfrac{dx}{dt} = -kx$
$x = x_oe^{-kt}$
When t = 25 yrs., x = (100% - 1.1%)xo = 0.989xo
$0.989x_o = x_oe^{-25k}$
$e^{-k} = 0.989^{1/25}$
Thus,
$x = 0.989^{t/25}x_o$
When x = 0.5xo
$0.5x_o = 0.989^{t/25}x_o$
$0.5^{25} = 0.989^t$
$t = \dfrac{25 \ln 0.5}{\ln 0.989}$
$t = 1566.65 ~ \text{yrs}$ answer