SAT Nonlinear Functions Question 60 | Free SAT Prep | Aniko

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Question 60·Hard·Nonlinear Functions

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The population of a certain bacterium in a laboratory culture is modeled by

P(t)=\frac{5,000}{1+9e^{-0.6t}},

where $P(t)$ is the number of bacteria and $t$ is the time, in hours, since the culture was started.

According to the model, approximately how many hours does it take for the population to increase from 1,000 bacteria to 4,000 bacteria?

For logistic or exponential growth questions on the SAT, translate the verbal description into equations by setting the given formula equal to the specified population values, then solve algebraically: clear the fraction, isolate the exponential term, and use natural logarithms to solve for the variable in the exponent. Always double-check what the question is actually asking—often it wants a time interval (difference of two times) rather than a single time—so find both relevant times and subtract, then round to match the nearest answer choice.

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Set up equations for the two population levels

Write two separate equations: one with $P(t)=1000$ and another with $P(t)=4000$ , both using the given formula for $P(t)$ . Think about solving each one for $t$ .

Isolate the exponential term

For each equation, clear the fraction by multiplying both sides, then move terms so that you get something like $e^{-0.6t} = \text{number}$ .

Use natural logarithms to solve for time

Once you have $e^{-0.6t}$ equal to a number, take the natural log (ln) of both sides to bring the exponent down and solve for $t$ .

Remember what the question is actually asking

You will get one time for 1,000 bacteria and another for 4,000. The question wants how long it takes to go from 1,000 to 4,000, so consider the difference between these two times, not just one of them.

Desmos Guide

Enter the population function

In Desmos, type P(t) = 5000/(1+9*exp(-0.6*t)). This defines the function modeling the bacteria population.

Find the time when the population is 1,000

Type P(t) = 1000 on a new line to create a horizontal line at 1000 and use the intersection tool (or click where the graphs meet) to read off the corresponding $t$ -value; this is the time when the population first reaches 1,000 bacteria.

Find the time when the population is 4,000

Similarly, type P(t) = 4000 to create a horizontal line at 4000 and find the intersection point with the curve $P(t)$ ; note the $t$ -value there as the time when the population reaches 4,000 bacteria.

Compute the time difference

In an expression line, type the larger $t$ -value minus the smaller one (for example, t2 - t1 using your recorded values) and let Desmos calculate the difference; this difference in hours is the value that matches one of the answer choices.

Step-by-step Explanation

Translate the question into equations for time

We are given

P(t)=\frac{5000}{1+9e^{-0.6t}},

and we want the time it takes to go from 1,000 to 4,000 bacteria.

That means:

Find $t_1$ such that $P(t_1)=1000$ .
Find $t_2$ such that $P(t_2)=4000$ .
Then compute the time difference $t_2 - t_1$ .

So the main work is solving two equations of the form $P(t)=\text{given number}$ for $t$ .

Solve the equation for 1,000 bacteria

Set $P(t)=1000$ and solve for $t$ :

1000 = \frac{5000}{1+9e^{-0.6t}}

Multiply both sides by $1+9e^{-0.6t}$ :

1000(1+9e^{-0.6t}) = 5000

Divide both sides by 1000:

1 + 9e^{-0.6t} = 5

Subtract 1 from both sides:

9e^{-0.6t} = 4

Divide by 9:

e^{-0.6t} = \frac{4}{9}

Now take the natural logarithm (ln) of both sides:

-0.6t = \ln\left(\frac{4}{9}\right)

t = \frac{\ln\left(\frac{4}{9}\right)}{-0.6}.

This value is the time $t_1$ when the population is 1,000 bacteria.

Approximate the time for 1,000 bacteria

Use approximate logarithms:

$\ln(4) \approx 1.386$
$\ln(9) \approx 2.197$

Then

\ln\left(\frac{4}{9}\right) = \ln(4) - \ln(9) \approx 1.386 - 2.197 = -0.811.

t_1 = \frac{-0.811}{-0.6} \approx 1.35 \text{ hours}.

Keep this value in mind; we will subtract it from the time for 4,000 bacteria later.

Solve the equation for 4,000 bacteria

Now set $P(t)=4000$ and solve for $t$ :

4000 = \frac{5000}{1+9e^{-0.6t}}

Multiply both sides by $1+9e^{-0.6t}$ :

4000(1+9e^{-0.6t}) = 5000

Divide both sides by 4000:

1 + 9e^{-0.6t} = \frac{5000}{4000} = \frac{5}{4}

Subtract 1 from both sides:

9e^{-0.6t} = \frac{5}{4} - 1 = \frac{1}{4}

Divide by 9:

e^{-0.6t} = \frac{1}{36}

Take the natural logarithm of both sides:

-0.6t = \ln\left(\frac{1}{36}\right) = -\ln(36)

t = \frac{-\ln\left(\frac{1}{36}\right)}{0.6} = \frac{\ln(36)}{0.6}.

This is the time $t_2$ when the population reaches 4,000 bacteria.

Approximate the time for 4,000 bacteria and find the difference

Approximate $t_2$ :

$\ln(36) \approx 3.584$

Then

t_2 = \frac{3.584}{0.6} \approx 5.97 \text{ hours}.

The time it takes to go from 1,000 to 4,000 bacteria is

t_2 - t_1 \approx 5.97 - 1.35 = 4.62 \text{ hours},

which rounds to $4.6$ hours. Therefore, the correct choice is C) 4.6.

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Step-by-step Explanation

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Step-by-step Explanation