Question

NO LINKS! PLEASE HELP ME!​

Answer 1

4860 bacteria

Explanation:

The given scenario can be modelled using an exponential function with base e.

`\boxed{\begin{minipage}{9 cm}\underline{General form of an Exponential Function with base $e$}\\\\$y=Ae^(kx)$\\\\where:\\\phantom{ww}$\bullet$ $A$ is the initial value ($y$-intercept). \\ \phantom{ww}$\bullet$ $e$ is Euler's number. \\ \phantom{ww}$\bullet$ $k$ is some constant.\\\end{minipage}}`

Define the variables:

• Let y be number of bacteria.
• Let x be the time (in hours)

Given the initial number of bacteria is 20, then A = 20.

`\implies y=20e^(kx)`

If the bacteria triples its number in 16 hours, then when x = 16:

`\implies y=3 * 20= 60`

Substitute x = 16 and y = 60 into the equation and solve for k:

`\begin{aligned}y&=20e^(kx)\\\implies 60&=20e^(16k)\\(60)/(20)&=e^(16k)\\3&=e^(16k)\\\ln 3&=\ln e^(16k)\\ \ln3&=16k \ln e\\\ln3&=16k\\k&=(1)/(16) \ln 3\end{aligned}`

Therefore, the equation that models the given scenario is:

`\large\boxed{y=20e^{\left((1)/(16)x \ln 3\right)}}`

To find how many bacteria there are after 80 hours, substitute x = 80 into the equation and solve for y:

`\implies y=20e^{\left((1)/(16)(80) \ln 3\right)}`

`\implies y=20e^(\left(5 \ln 3\right))`

`\implies y=20e^(\left(\ln 3^5\right))`

`\implies y=20e^(\left(\ln 243\right))`

`\implies y=20 \cdot 243`

`\implies y=4860`

Therefore, there are 4860 bacteria after 80 hours.

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Please note that I have solved this question so that you can calculate the number of bacteria for any number of hours, part hours, etc.

However, there is a much simpler way of solving this particular question.

If the number of bacteria triples every 16 hours, then it will triple 5 times within 80 hours since 80 ÷ 16 = 5.

Therefore:

• Hour 0: 20
• Hour 16: 20 × 3 = 60
• Hour 32: 60 × 3 = 180
• Hour 48: 180 × 3 = 540
• Hour 64: 540 × 3 = 1620
• Hour 80: 1620 × 3 = 4860

As you can see, the result is the same as the previous method.