Answer:
Explanation:
The given equation assumes a 12 cubic inch initial volume. The question wants to know when, in minutes, the volume, y, reaches 36 in^3:
y = 12*(2)^(t/60), where t is time in minutes
Therefore:
36 = 12*(2)^(t/60)
3 = (2)^(t/60)
log(3) =(60/t) log(2)
t = 60log(3)/log(2)
t = 95.1 minutes