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Difficulty: MediumExponential Functions and Equations

A scientist models the population of a bacteria culture using the function P(t)=P0⋅btP(t) = P_0 \cdot b^t, where P(t)P(t) is the population tt hours after the start of the experiment, P0P_0 is the initial population, and bb is a constant. The table below shows the population at two different times:

tt (hours)P(t)P(t)
221,8001,800
5548,60048,600

If the population of the bacteria culture is 1,312,2001,312,200 after kk hours, what is the value of kk?

  1. A
    4
  2. 8Answer
  3. C
    9
  4. D
    14

Answer

8
The correct value is 8. By setting up the ratio of the population at t=5t = 5 to t=2t = 2, we find that P(5)P(2)=b3=48,6001,800=27\frac{P(5)}{P(2)} = b^3 = \frac{48,600}{1,800} = 27, which yields an hourly growth factor of b=3b = 3. Using P(2)=P0⋅32=1,800P(2) = P_0 \cdot 3^2 = 1,800, we determine the initial population P0P_0 is 200. To find the hour kk when the population is 1,312,2001,312,200, we solve 200⋅3k=1,312,200200 \cdot 3^k = 1,312,200, which simplifies to 3k=6,5613^k = 6,561. Since 38=6,5613^8 = 6,561, we find k=8k = 8. Alternatively, we can calculate P(k)P(5)=1,312,20048,600=27=33\frac{P(k)}{P(5)} = \frac{1,312,200}{48,600} = 27 = 3^3, meaning the population triples 3 more times after t=5t = 5, giving k=5+3=8k = 5 + 3 = 8.

Step-by-Step Solution

1
Set up the ratio of the population at t=5t = 5 to the population at t=2t = 2 to find the growth factor bb.
P(5)P(2)=P0⋅b5P0⋅b2=b3=48,6001,800=27\frac{P(5)}{P(2)} = \frac{P_0 \cdot b^5}{P_0 \cdot b^2} = b^3 = \frac{48,600}{1,800} = 27
Dividing the function values eliminates the initial population P0P_0 and isolates the base bb.
2
Solve for the growth factor bb.
b=3b = 3
Since b3=27b^3 = 27, taking the cube root of both sides gives b=3b = 3.
3
Find the initial population P0P_0 using P(2)=1,800P(2) = 1,800 and b=3b = 3.
1,800=P0⋅32⇒1,800=9P0⇒P0=2001,800 = P_0 \cdot 3^2 \Rightarrow 1,800 = 9P_0 \Rightarrow P_0 = 200
Substituting the known values into the function formula allows us to solve for P0P_0.
4
Set up the equation for the population after kk hours and solve for kk.
1,312,200=200⋅3k⇒3k=1,312,200200=6,5611,312,200 = 200 \cdot 3^k \Rightarrow 3^k = \frac{1,312,200}{200} = 6,561
Dividing both sides by 200 isolates the exponential term 3k3^k.
5
Find the exponent kk by expressing 6,561 as a power of 3.
3k=38⇒k=83^k = 3^8 \Rightarrow k = 8
Since 38=6,5613^8 = 6,561, the exponents must be equal, so k=8k = 8.

Key Concept

Determining parameters of an exponential growth function from given data points and using the function to solve for an unknown time variable.

Alternative Method

Instead of solving for the initial population P0P_0, we can compare the target population of 1,312,2001,312,200 to the population at t=5t = 5. Since the population triples every hour (b=3b = 3), the ratio of the population at t=kt = k to the population at t=5t = 5 is 3k−53^{k-5}. We have 1,312,20048,600=27\frac{1,312,200}{48,600} = 27, which is 333^3. Therefore, 3k−5=333^{k-5} = 3^3, which directly gives k−5=3k - 5 = 3, or k=8k = 8.
Estimated Time:1m 15s
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Topics

Exponential Functions and Equations

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