Question

no calculator is allowed on this question.
the initial number of a bacteria in a colony is 500. if the number of bacterial doubles every day, which function represents the number of bacteria after ( d ) days?
select one answer
a ( b(d) = 250(2)^d )
b ( b(d) = 500(d)^2 )
c ( b(d) = (500d)^2 )
d ( b(d) = 500(2)^d )

Explanation:

Step1: Recall exponential growth formula

The general formula for exponential growth is \( B(d)=a(r)^d \), where \( a \) is the initial amount, \( r \) is the growth rate, and \( d \) is the time (in days here).

Step2: Identify initial amount and growth rate

The initial number of bacteria \( a = 500 \). The number doubles every day, so the growth rate \( r = 2 \).

Step3: Substitute into the formula

Substituting \( a = 500 \) and \( r = 2 \) into the exponential growth formula, we get \( B(d)=500(2)^d \)? Wait, no, looking at the options, option D is \( B(d) = 500(2)^d \)? Wait, let's check the options again. Wait, the options:

A: \( B(d)=250(2)^d \) (initial is 250, wrong)

B: \( B(d)=500(d)^2 \) (quadratic, not exponential growth for doubling, wrong)

C: \( B(d)=(500d)^2 \) (quadratic, wrong)

D: \( B(d)=500(2)^d \) (initial 500, doubling every day, so exponential growth with base 2, correct)

Wait, maybe I misread the initial. The problem says initial number is 500, doubles every day. So the formula should be \( B(d)=500\times2^d \), which is option D.

Answer:

D. \( B(d) = 500(2)^d \)