To solve the recurrence relation T(n) = T(n/2) + 2^n using the master theorem, we need to compare the function 2^n with n^(log_b a) = n^(log_2 1) = n^0 = 1.
Since 2^n grows much faster than n^0, we can apply case 3 of the master theorem, which states that if f(n) = 2^n, and if there exists a constant ε > 0 such that f(n) = Ω(n^(log_b a + ε)), then T(n) = Θ(f(n)).
In our case, since 2^n = Ω(n^(0 + ε)) for any ε > 0, we can conclude that T(n) = Θ(2^n).
Therefore, the solution to the recurrence relation T(n) = T(n/2) + 2^n is T(n) = Θ(2^n).