Hi, I'm really lost..

Let W(t) be a standard Brownian Motion,
$\displaystyle X_t = t + 2^{-t}W(4^t), t>=0 $

Find the covariance function of the process $\displaystyle X_t $

here is my working out, but im not sure if im doing it right, or what i should do next:
$\displaystyle cov (X_t,X_s) $
$\displaystyle = cov (t,X_s) + cov (2^{-t}W(4^t), X_s) $
$\displaystyle = t cov(1,X_s) + 2^{-t}cov(W(4^t), X_s) $
$\displaystyle = t cov (1,X_s) + ts cov(1,X_s) + 2^{-t} cov(W(4^t),X_s) + 2^{-t} scov(W(4^t),1) $
$\displaystyle = 2^{-t} cov(W(4^t),X_s) $

and now i don't know what to do, please help

here is the answer:
$\displaystyle R(t,s) = Cov(X_t,X_s) = 2^{-|t-s|} $