Master Theorem: How can $ 2^m *2^.5 = m/2?$

Question

I'm reading about solving this recurrence relation using Master theorem:

$T(n)=2T(√n)+Θ(log n)$

In the solution m = log n , making the above:

$T(2^m) = 2T(2^m *2^.5)+Θ(m)$

Then changing S(m) to $T(2^m)$:

$S(m)=2S(m/2)+Θ(m)$

My question is: How come $2^m *2^.5 = m/2$ ?

Answer 1

Your mistake is when you substitute $m$ for $\log _2 n$: this implies that $n = 2^m$, so $\sqrt n = (2^m)^{\frac 1 2} = 2 ^ {\frac m 2}$, not $2^m 2^{\frac 1 2}$ as you have written.