$x = x_oe^{-kt}$
When t = 38 hr, x = 0.5xo
$0.5x_o = x_oe^{-38k}$
$e^{-k} = 0.5^{1/38}$
Hence,
$x = x_o(0.5^{t/38})$
When 90% are dissipated, x = 0.1xo
$0.1x_o = x_o(0.5^{t/38})$
$0.1 = 0.5^{t/38}$
$0.1^{38} = 0.5^t$
$t = \dfrac{38 \ln 0.1}{\ln 0.5}$
$t = 126.23 ~ \text{hrs}$ answer