A scientist records the temperature, (in °C), of a liquid as it cools in a laboratory that is kept at a constant . The data show that at minutes, , and at minutes, . For , the relationship between time (in minutes) and temperature is well modeled by the exponential function where and are constants. Based on the two data points, which of the following is the best estimate of the value of ?

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SAT Two-Variable Data Question 57 | Free SAT Prep | Aniko

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Question 57·Hard·Two-Variable Data: Models and Scatterplots

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A scientist records the temperature, $T$ (in °C), of a liquid as it cools in a laboratory that is kept at a constant $20\,^{\circ}\!\text{C}$ . The data show that

at $t = 0$ minutes, $T = 92\,^{\circ}\!\text{C}$ , and
at $t = 5$ minutes, $T = 65\,^{\circ}\!\text{C}$ .

For $t>0$ , the relationship between time $t$ (in minutes) and temperature $T$ is well modeled by the exponential function

T = 20 + a\,b^{t},\quad 0<b<1,

where $a$ and $b$ are constants.

Based on the two data points, which of the following is the best estimate of the value of $b$ ?

For exponential models with a horizontal asymptote, first subtract the asymptote (here, 20°C) so you are working with the excess amount above that baseline. Use the $t=0$ data point to solve for the initial amount $a$ , then plug in the second data point to find an equation for $b^t$ (here, $b^5$ ). Solve for that power, and if the exponent is not 1, use roots or test the answer choices by raising each option to the given power and seeing which one matches the computed value. This avoids messy algebra and lets you quickly identify the correct base from the choices.

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Start with the point at $t = 0$

Plug $t = 0$ and $T = 92$ into $T = 20 + a b^t$ . Remember that $b^0$ has a simple value. Use this to solve for $a$ first.

Use the second data point to relate to $b$

After you find $a$ , substitute $t = 5$ and $T = 65$ into the model. Rearrange the equation so that you isolate $b^5$ on one side.

Focus on the temperature above room temperature

Notice that $T - 20$ equals $a b^t$ . For $t=0$ and $t=5$ , compare how much higher than $20^{\circ}\!\text{C}$ the liquid is. That ratio tells you $b^5$ .

Compare answer choices using powers, not guessing

Once you know the value of $b^5$ , raise each answer choice to the 5th power (mentally or with a calculator) and see which one is closest to that value.

Desmos Guide

Compute the value of $b^5$ from the data

In Desmos, type 5/8 and note the decimal value (this is the value of $b^5$ that comes from the equation).

Compare each answer choice to the required $b^5$ value

Either:

Type (5/8)^(1/5) to directly see the decimal value of $b$ , then compare that number to the answer choices, or
Separately type 0.6^5, 0.8^5, 0.91^5, and 0.95^5 and compare each result to 5/8. The choice whose 5th power is closest to 5/8 is the correct value of $b$ .

Step-by-step Explanation

Use the initial temperature to find $a$

We are given the model

T = 20 + a b^{t}

and the data point $t=0$ , $T=92$ .

Plug $t=0$ and $T=92$ into the equation:

92 = 20 + a b^{0}

Since $b^{0} = 1$ ,

92 = 20 + a \Rightarrow a = 92 - 20 = 72.

So the model becomes

T = 20 + 72 b^{t}.

Use the second data point to solve for $b^5$

Now use the point $t=5$ , $T=65$ in the updated model $T = 20 + 72 b^{t}$ .

Plug in $t=5$ and $T=65$ :

65 = 20 + 72 b^{5}.

Subtract 20 from both sides:

\begin{aligned} 65 - 20 &= 72 b^{5} \\ 45 &= 72 b^{5}. \end{aligned}

Divide both sides by 72:

b^{5} = \frac{45}{72}.

Simplify the fraction by dividing numerator and denominator by 9:

b^{5} = \frac{5}{8} = 0.625.

So we need $b$ such that $b^{5} \approx 0.625$ . Now compare the answer choices.

Test the answer choices by raising them to the 5th power

Raise each answer choice to the 5th power and compare it to $0.625$ (or $5/8$ ):

$0.60^{5}$ is very small (around $0.08$ ), far below $0.625$ .
$0.80^{5}$ is about $0.33$ , still much less than $0.625$ .
$0.95^{5}$ is about $0.77$ , which is larger than $0.625$ .
$0.91^{5}$ is approximately $0.624$ , which is extremely close to $0.625$ .

Therefore, the best estimate for $b$ is 0.91.

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Step-by-step Explanation

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