SAT Nonlinear Functions Question 12 | Free SAT Prep | Aniko

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Question 12·Hard·Nonlinear Functions

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Growth of a Culture of Bacteria

Day	Number of bacteria per milliliter at end of day
1	$2.5 \times 10^5$
2	$5.0 \times 10^5$
3	$1.0 \times 10^6$

A culture of bacteria is growing at an exponential rate, as shown in the table above. At this rate, on which day would the number of bacteria per milliliter reach $5.12 \times 10^8$ ?

For exponential growth questions from a table, first decide if the pattern is additive (linear) or multiplicative (exponential) by checking if you add or multiply by a constant to go from one row to the next. Once you see a constant multiplier, write a formula like starting_value × (growth_factor)^(days_since_start), then plug in the target amount and solve for the exponent. On the SAT, growth factors are usually small integers (like 2 or 3), so look for familiar powers (such as $2^{10}=1024$ ) to solve quickly without using logarithms.

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Identify the pattern in the table

Compare the bacteria counts from day to day. Are you adding the same amount each day, or multiplying by the same factor?

Write a formula for day d

Use the day 1 value as a starting point. If the number multiplies by the same factor each day, how can you write the amount on day $d$ in terms of that factor and $d-1$ ?

Set up and simplify the equation

Set your exponential formula equal to $5.12\times 10^8$ , then divide to isolate the power term. What number do you get, and can you express it as a power of 2?

Relate the exponent to the day number

Once you match the number on the right to a power of 2, set the exponents equal and solve for the day $d$ .

Desmos Guide

Enter the exponential model

In Desmos, type y = 2.5*10^5 * 2^(x-1) to represent the number of bacteria $N(d)$ as a function of the day $x$ .

Enter the target value

On a new line, type y = 5.12*10^8 to create a horizontal line showing the desired bacteria count.

Find the intersection

Use Desmos’s intersection tool (tap on the point where the curve and horizontal line meet, or click the intersection icon) and read the $x$ -coordinate at that point; that $x$ -value is the day when the bacteria count reaches $5.12*10^8$ .

Step-by-step Explanation

Find the growth factor from the table

Look at how the bacteria count changes each day:

Day 1: $2.5\times 10^5$
Day 2: $5.0\times 10^5$ (this is $2$ times $2.5\times 10^5$ )
Day 3: $1.0\times 10^6$ (this is $2$ times $5.0\times 10^5$ )

So the number of bacteria is doubling each day. The growth factor is $2$ per day.

Write an exponential model for day d

Let $d$ be the day number and $N(d)$ be the number of bacteria per milliliter on day $d$ .

On day 1, $N(1)=2.5\times 10^5$ . Each day, we multiply by $2$ again. After $d-1$ days of doubling, we get:

N(d) = 2.5\times 10^5 \times 2^{d-1}

This formula matches the table: for $d=1,2,3$ it gives the given values.

Set up an equation using the target bacteria count

We are told the culture reaches $5.12\times 10^8$ bacteria per milliliter. Set $N(d)$ equal to this value:

2.5\times 10^5 \times 2^{d-1} = 5.12\times 10^8

Now divide both sides by $2.5\times 10^5$ to isolate the exponential part:

2^{d-1} = \frac{5.12\times 10^8}{2.5\times 10^5}

Simplify the right side:

2^{d-1} = \frac{5.12}{2.5} \times 10^{8-5} = 2.048 \times 10^3 = 2048

Express 2048 as a power of 2 and solve for the day

Now we have

2^{d-1} = 2048

Recognize that $2048$ is a power of $2$ :

$2^{10} = 1024$
$2^{11} = 2048$

So $2048 = 2^{11}$ , which gives

2^{d-1} = 2^{11} \Rightarrow d-1 = 11 \Rightarrow d = 12

So the number of bacteria reaches $5.12\times 10^8$ on Day 12.

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Step-by-step Explanation

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Step-by-step Explanation