A laboratory is monitoring the decay of a radioactive substance. The function above models the mass , in milligrams, of the substance days after the start of the experiment. Which of the following best describes the meaning of the factor in this context?

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SAT Nonlinear Functions Question 182 | Free SAT Prep | Aniko

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Question 182·Medium·Nonlinear Functions

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p(t)=50(0.6)^{\frac{t}{5}}

A laboratory is monitoring the decay of a radioactive substance. The function above models the mass $p(t)$ , in milligrams, of the substance $t$ days after the start of the experiment.

Which of the following best describes the meaning of the factor $0.6$ in this context?

For exponential word problems, first match the function to the pattern $a \cdot b^{t/k}$ : identify $a$ as the initial amount and $b$ as the growth/decay factor applied every $k$ time units (because the exponent increases by 1 every $k$ units of time). Plug in a couple of easy time values (like $t=0$ and $t=k$ ) to see how the quantity changes, then choose the option whose wording matches that repeated multiplication factor, paying close attention to "each day" vs. "every 5 days" and to whether the factor describes what remains or what is lost.

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Focus on the structure of the function

Look at $p(t)=50(0.6)^{t/5}$ . Which number represents the starting amount, and which number is being raised to a power?

Think about what t/5 means

Ask yourself: If the exponent were just $t$ , the factor would apply every 1 day. But here it is $t/5$ . How often does the exponent go up by 1 when time increases?

Compare the function values at different times

Try plugging in $t=0$ and $t=5$ . How does $p(5)$ compare to $p(0)$ ? Is the mass being multiplied or reduced by a certain fraction over that interval?

Desmos Guide

Enter the function and create a table

Type p(t)=50*(0.6)^(t/5) into Desmos. Then add a table with $t$ -values like 0, 5, 10 in the first column and use p(t) in the second column so Desmos fills in the masses.

Compare values over 5-day intervals

Look at the values of $p(0)$ and $p(5)$ . Find the ratio $p(5) / p(0)$ (you can also type p(5)/p(0) in Desmos). Notice this constant ratio over each 5-day increase in $t$ , and interpret what that factor tells you about how much of the substance remains over each 5-day period.

Step-by-step Explanation

Identify the parts of the exponential model

The function is

p(t)=50(0.6)^{t/5}

In an exponential model of the form $a \cdot b^{\text{(time expression)}}$ :

$a$ is the initial amount (here, 50 milligrams at $t=0$ ).
$b$ is the multiplication factor applied each time the exponent increases by 1.

Interpret the exponent t/5

The exponent is $\frac{t}{5}$ , which counts how many 5-day periods have passed.

When $t=0$ , the exponent is $0/5 = 0$ (no 5-day periods yet).
When $t=5$ , the exponent is $5/5 = 1$ (one 5-day period).
When $t=10$ , the exponent is $10/5 = 2$ (two 5-day periods).

Each time $t$ increases by 5 days, the exponent increases by 1, so the mass is multiplied by $0.6$ once every 5 days.

See what multiplying by 0.6 every 5 days means and match the choice

Compute the mass at key times:

At $t=0$ :

p(0) = 50(0.6)^{0} = 50 \cdot 1 = 50

At $t=5$ :

p(5) = 50(0.6)^{1} = 50 \cdot 0.6

So after 5 days, the mass is $0.6$ times the initial mass, meaning 60% of the mass is still there (40% has been lost).

This pattern repeats every 5 days: each 5-day interval, the mass is multiplied by $0.6$ , so 60% of the substance’s mass remains every 5 days after the experiment begins. That matches choice C.

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Step-by-step Explanation

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