Question

A sample of bacteria is decaying according to a half-life model. If the sample begins with 500 bacteria, and after 14 minutes there are 100 bacteria, after how many minutes will there be 50 bacteria remaining?

When solving this problem, round the value of k to four decimal places and round your final answer to the nearest whole number.

Answer 1

20 minutes

Explanation:

`\boxed{\begin{minipage}{9 cm}\underline{General form of an Exponential Function with base e}\\\\$f(t)=Ae^(kt)$\\\\where:\\\phantom{ww}$\bullet$ $A$ is the initial value ($y$-intercept). \\ \phantom{ww}$\bullet$ $e$ is the base (Euler's number).\\ \phantom{ww}$\bullet$ $k$ is some constant.\\\end{minipage}}`

Given values:

• A = 500
• f(14) = 100

Substitute these values into the formula and solve for k:

`\implies 100=500e^(14k)`

`\implies (100)/(500)=e^(14k)`

`\implies e^(14k)=0.2`

`\implies \ln e^(14k)=\ln 0.2`

`\implies 14k \ln e=\ln 0.2`

`\implies 14k=\ln 0.2`

`\implies k=(1)/(14)\ln 0.2`

`\implies k=-0.1150\;\; \sf (4\; d.p.)`

Therefore, the function that models the scenario is:

`\boxed{f(t)=500e^(-0.1150t)}`

To find how many minutes it takes for the bacteria to be 50, substitute f(t)=50 into the equation and solve for t:

`\implies 500e^(-0.1150t)=50`

`\implies e^(-0.1150t)=(50)/(500)`

`\implies e^(-0.1150t)=0.1`

`\implies \ln e^(-0.1150t)=\ln 0.1`

`\implies -0.1150t\ln e=\ln 0.1`

`\implies -0.1150t=\ln 0.1`

`\implies t=(\ln 0.1)/(-0.1150)`

`\implies t=20 \; \sf minutes\;\; (nearest\;whole\;number)`

Therefore, there will be 50 bacteria remaining after 20 minutes.

Answer 2

20 minutes

Explanation:

If a population of 500 bacteria decays to 100 in 14 minutes, you want to know how long it takes for the population to decline to 50.

Exponential equation

The equation modeling the population can be written ...

p(t) = (initial population) · (growth factor)^(t/(growth period))

where "growth period" is the period associated with the "growth factor".

Application

In this problem, we have ...

• initial population = 500
• growth factor = 100/500 = 1/5
• growth period = 14 minutes

So, the population equation can be written ...

p(t) = 500(1/5)^(t/14)

We want to find the value of t that brings the population to 50:

50 = 500(1/5)^(t/14)

50/500 = (1/5)^(t/14) . . . . divide by 500

log(1/10) = (t/14)·log(1/5) . . . . take logarithms

t = 14·log(1/10)/log(1/5) . . . . divide by the coefficient of t

t ≈ 20

After 20 minutes there will be 50 bacteria remaining.

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Additional comment

You might be expected to work this using the exponential equation ...

p(t) = 500e^(-kt)

The value of k will be -ln(1/5)/14 ≈ 1.60944/14 ≈ 0.1150 (rounded to 4 dp)

Then you're solving ...

ln(50/500) = -0.1150t

t = ln(0.1)/-0.1150 ≈ 20.029 ≈ 20

We like the exponential form show above, because it does not involve rounding of any of the values in the equation. They all come directly from numbers in the problem.