Please stuck

Question

A species of bacteria has an hourly growth rate of 5 and an initial population of 12. Which exponential function would represent the growth rate after t hours?
a)f(x)=5(12)^-t
b)f(x)=5(12)^t
c)f(x)=12(5)^t
d)f(x)=12(5)^-t

Answer 1

Lucy, let's look at what we have. The initial population stays the same, so we are down to c or d. After year one, the growth is 12(5). After year two, the growth is 12(5)(5), or 12(5)².

 

The answer is c. With d, after the second year, the growth would be 12(5)^1/2, or 12 times the square root of 5. Year three would be 12 times the cube root of 5, so your population would be shrinking!

Answer 2

A = a₀(r)^t, where A = result, a₀ = initial amount, r = rater, t = time.

 

c) is the only one that fits this form.

Answer 3

If this is an exponential growth problem then

 

df/dt = kf

 

k = 5 hours^-1

 

The general solution of this type of equation is the exponential function

 

f(t) = f(0)e^kt

 

Given:  f(0) = 12, k = 5

 

f(t) = 12e^5t

 

Then the growth rate after t hours is

 

df/dt = k[f(0)e^kt] = 5 * (12e^5t)

 

But this is not among the answers in the list.