Please help! A bacteria culture starts with 500 bacteria and doubles every 3.5 hours. Find an expression for the number of bacteria after t hours.
a.550e^35t
b.500=Qoe^(ln2)/(3.5)(t)
c.500e^(ln2)/(3.5)(t)
d.2e^3.5t

Question

Please help! A bacteria culture starts with 500 bacteria and doubles every 3.5 hours. Find an expression for the number of bacteria after t hours. 
a.550e^35t
b.500=Qoe^(ln2)/(3.5)(t) 
c.500e^(ln2)/(3.5)(t)
d.2e^3.5t

anonymous · Answer

at t=0        N = 500
after 3.5   N = 500 * 2
after 7.0   N = 500 * 2 * 2
after 10.5 N = 500 * 2 * 2 * 2

after 3.5*k hours,   $N_k=500(2^k)$