Question

How do I solve this exponential equation problem?

Answer 1

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(1) 750

(2) 1755
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(3) 16.309

Explanation:

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The exponential growth model is:

`color(red)(A(t) = A_0 e^(kt)`, where

`color(red)(A(t)` represents the amount of bacteria after the amount of time `color(red)(t)`

`color(red)(A_0` represents the amount of bacteria at the beginning of the exponential growth.

`color(red)(k` is a constant value, it varies from problem to problem.

We are given the exponential equation `color(blue)(A(t) = 750e^(0.0425t)`

Hence,

`A_0 = 750`

`k=0.0425`

Now we are ready to answer the questions:

1. What is the initial number of bacteria ?

   Answer: 750
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   2. What is the number of bacteria after 20 minutes ?

   Answer: `A(20)=750e^(0.0425*20)`
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   using the calculator,

`A(20) ~~ 1754.73514`

`A(20) ~~ 1755` ( Rounded to the nearest whole number)

Hence, the number of bacteria after 20 minutes is 1755.
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3. How long will it take for the number of bacteria to double ?

Find the mount of time (in minutes) it takes for the number of bacteria to be 1500.

`A(t) = 750e^(0.0425*t)`

For this question, we know that `A(t) = 1500`

Hence, we must solve for `t` using the exponential equation:

`1500=750e^(0.0425*t)`

To find the value of `t`, divide both sides of the equation by `750`

`1500/750=(750e^(0.0425*t))/750`

`cancel (1500)^color(red)(2)/cancel (750)=(cancel (750)e^(0.0425*t))/cancel(750`

`2=e^(0.0425*t)`

To simplify, take the natural log of both sides of the equation:

`ln 2 = ln e^(0.0425*t)`

Move the exponent to the front and simplify.

`ln 2 =0.0425*t* (ln e)`

Note that `color(blue)(ln e=1`

Hence, we get

`ln 2 =0.0425*t* 1`

`ln 2 =0.0425*t`

To solve for `t`, divide both sides by `0.0425`

We get,

`(ln 2)/0.0425 =(0.0425*t)/0.0425`

`(ln 2)/0.0425 =(cancel 0.0425*t)/cancel 0.0425`

Swap sides:

`t=(ln 2/0.0425)`

`t~~ 16.30934542` (using a calculator)

`t~~ 16.309` (accurate to 3 decimal places)

Hence, it takes 16.309 minutes approximately for the bacteria to double.

Hope you find the solution useful.