Question

Scientists have collected a sample of the bacteria responsible for an illness and determined the equation below:

B = 100e^(0.592t)
B is the number of bacteria after t hours.

a. Describe the graph of the exponential equation. Is it growth or decay? Does it have any asymptotes?
b. How much bacteria will there be in 1, 5, 10, and 24 hours?
c. We need to be able to find how long the bacteria have been present given the amount of bacteria we find. To do this we need to solve the equation for time using logs. What is the log equation to calculate time?
d. There is found to be 3500 bacteria; how long has the bacteria been growing?

Answer 1

a.
it is grouth because the exponent is positive
the vertical assemtote is y=0

b.
1 hour: 100e^(0.592(1))=100e^(0.592)≈180.76 bacteria
5 hours: 100e^(0.592(5))≈1929.80 bacteria
10 hours: 100e^(0.592(10))≈37241.17 bacteria
24 hours: 100e^(0.592(24))≈148066223.24 bacteria


c. solve for t
I will solve using both ln and log₁₀ in case your teacher wants one or the other

so
log₁₀ first
B=100e^(0.592t)
divide both sides by 100
B/100=e^(0.592t)
take log₁₀ both sides
log₁₀(B/100)=log₁₀(e^0.592t)
log₁₀(B/100)=0.0592tlog₁₀(e)
divide both sides by 0.592log₁₀(e)

`(log_(10)((B)/(100)))/(0.592log_(10)(e))=t`

now using ln
B=100e^(0.592t)
divide by 100 both sides
B/100=e^(0.592t)
take ln of both sides
ln(B/100)=0.592tln(e)
ln(B/100)=0.592t
divide both sides by 0.592
(ln(B/100))/0.592=t



d.
using our equation (I'll use ln)
(ln(B/100))/0.592=t
(ln(3500/100))/0.592=t
(ln(35))/0.592=t
6.00566=t
so about 6 hours