$\dfrac{dx}{dt} = -kx$

$x = x_oe^{-kt}$

When t = 25 yrs., x = (100% - 1.1%)x_{o} = 0.989x_{o}

$0.989x_o = x_oe^{-25k}$

$e^{-k} = 0.989^{1/25}$

Thus,

$x = 0.989^{t/25}x_o$

When x = 0.5x_{o}

$0.5x_o = 0.989^{t/25}x_o$

$0.5^{25} = 0.989^t$

$t = \dfrac{25 \ln 0.5}{\ln 0.989}$

$t = 1566.65 ~ \text{yrs}$ *answer*