$T(n) = T(n/2) + 1$

$=T(n/4) + 2$

$= T(n/8) + 3$

$\vdots$

$=T(n/{2^k}) + k.$

Recurrence stops when $2^k >= n$.

When $2^k = n,k = \lg n$

So, $T(n) = T(1) + \lg n \\= 1 + \lg n$

PS: Unless explicitly asked for asymptotic bound, we should give exact answers for solutions of recurrence equations.