Okay, let's solve this step-by-step:
* Original amount: 64 g
* Amount after t hours: 2 g
* Half-life: 1/2 hours
* To find t, we set up the half-life equation: N = N0 * (1/2)^(t/T1/2)
* Plug in the values: 2 = 64 * (1/2)^(t/0.5)
* Take the log of both sides: log(2) = log(64) - (t/0.5)log(2)
* Solve for t: t = 1 hour
Therefore, the time for the amount to reduce to 2