Surely you can just plug in your numbers.
#1: N(4) = 1000*e^.04
#2: 1000*e^.01t = 1700
#3 e^.01t = 2
1.What is the population after 4 hours?
2.When will the number of bacteria reach 1700?
3. When will the number of bacteria double?
Please help and show all work
#1: N(4) = 1000*e^.04
#2: 1000*e^.01t = 1700
#3 e^.01t = 2
1. To find the population after 4 hours, substitute t = 4 into the equation N(t) = 1000e^(0.01t):
N(4) = 1000e^(0.01 * 4)
N(4) = 1000e^(0.04)
Using a calculator, we can find that N(4) ≈ 1105.17 (rounded to 2 decimal places).
Therefore, the population after 4 hours is approximately 1,105.17 bacteria.
2. To find when the number of bacteria reaches 1700, we can set N(t) equal to 1700 and solve for t.
1700 = 1000e^(0.01t)
Divide both sides of the equation by 1000:
1.7 = e^(0.01t)
Now, take the natural logarithm (ln) of both sides to isolate the exponent:
ln(1.7) = ln(e^(0.01t))
Using the property ln(e^x) = x, we can simplify further:
ln(1.7) = 0.01t
Now, divide both sides by 0.01 to solve for t:
t = ln(1.7) / 0.01
Using a calculator, we can find that t ≈ 77.35 (rounded to 2 decimal places).
Therefore, the number of bacteria will reach 1700 after approximately 77.35 hours.
3. To find when the number of bacteria doubles, we can set N(t) equal to 2N(0), where N(0) is the initial population.
2N(0) = 1000e^(0.01t)
Divide both sides of the equation by 1000:
2 = e^(0.01t)
Take the natural logarithm of both sides to isolate the exponent:
ln(2) = ln(e^(0.01t))
Using the property ln(e^x) = x, we have:
ln(2) = 0.01t
Now, divide both sides by 0.01 to solve for t:
t = ln(2) / 0.01
Using a calculator, we can find that t ≈ 69.31 (rounded to 2 decimal places).
Therefore, the number of bacteria will double after approximately 69.31 hours.
Question 1: What is the population after 4 hours?
To find the population after 4 hours, we need to substitute the value of t=4 into the given function N(t) = 1000e^0.01t.
N(4) = 1000e^0.01(4)
Now, we can calculate this value using a basic calculator or a scientific calculator:
N(4) = 1000e^0.04 ≈ 1000 * 1.04 ≈ 1040
So, the population after 4 hours is approximately 1040 bacteria.
Question 2: When will the number of bacteria reach 1700?
To find the time when the number of bacteria reaches 1700, we need to solve the equation N(t) = 1700.
Substituting N(t) = 1700 into the original equation, we get:
1700 = 1000e^0.01t
To solve for t, we need to isolate the exponential term:
e^0.01t = 1700 / 1000
e^0.01t = 1.7
Now, take the natural logarithm (ln) of both sides to remove the exponential term:
0.01t = ln(1.7)
Finally, solve for t by dividing both sides by 0.01:
t = ln(1.7) / 0.01 ≈ 38.97
So, the number of bacteria will reach 1700 after approximately 38.97 hours.
Question 3: When will the number of bacteria double?
To find when the number of bacteria doubles, we need to find the time when N(t) is twice its initial value, which is 1000.
Substituting N(t) = 2 * 1000 = 2000 into the original equation, we get:
2000 = 1000e^0.01t
Dividing both sides of the equation by 1000, we get:
2 = e^0.01t
Again, take the natural logarithm (ln) of both sides:
ln(2) = 0.01t
Lastly, solve for t by dividing both sides by 0.01:
t = ln(2) / 0.01
Using a calculator, ln(2) ≈ 0.693:
t ≈ 0.693 / 0.01 ≈ 69.3
So, the number of bacteria will double after approximately 69.3 hours.
I hope this helps! Let me know if you have any further questions.