$x = x_oe^{-kt}$

When t = 38 hr, x = 0.5x_{o}

$0.5x_o = x_oe^{-38k}$

$e^{-k} = 0.5^{1/38}$

Hence,

$x = x_o(0.5^{t/38})$

When 90% are dissipated, x = 0.1x_{o}

$0.1x_o = x_o(0.5^{t/38})$

$0.1 = 0.5^{t/38}$

$0.1^{38} = 0.5^t$

$t = \dfrac{38 \ln 0.1}{\ln 0.5}$

$t = 126.23 ~ \text{hrs}$ *answer*