$T(n) = T(n/2) + 1$
$=T(n/4) + 2$
$= T(n/8) + 3$
$\vdots$
$=T(n/{2^k}) + k.$
Recurrence stops when $2^k \geq n$.
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.