SAT Functions as Models

To work with functions that model real-life situations on the SAT, you need to know:

• the Rules of Exponents
• how to translate words into math
• the definition of an exponential function

For any real number $x$, the exponential function $f(x)$ with base $a$ is defined as

$$f(x)=a^{x},\quad a>0,\ \ a\neq 1$$

• the definition of exponential growth and decay

Exponential growth (decay) occurs when the output of an exponential function increases (decreases) by a constant rate for a constant increase of its input.

Growth: $\qquad f(x)=ba^{x}, \quad a>1, \ \ b > 0$

Decay: $\qquad f(x)=ba^{x}, \quad 0< a < 1, \ \ b > 0$

$b$ denotes the initial value of $f(x)$, and the base $a$ is called the growth (decay) factor.

The relationship between the Celsius and Fahrenheit scales for temperature is given by the formula $F=\frac{9}{5}C+32,$ where $F$ is the temperature in degrees Fahrenheit and $C$ is the temperature in degrees Celsius. If the boiling point of curium is $3110^{\circ}$C, what is it in degrees Fahrenheit?

(A) $\ \ 622$
(B) $\ \ 3110$
(C) $\ \ 5598$
(D) $\ \ 5630$
(E) $\ \ 6700$

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Correct Answer: D

Solution:

Tip: Use a calculator.
Tip: Follow order of operations.
We're given that $C=3110^{\circ}$C. To solve for $F$, we substitute $C$ with $3110$ in the given formula.

$F=\frac{9}{5}C+32=\frac{9}{5}\cdot 3110+32= 9 \cdot 622 +32 = 5598+32 = 5630$ degrees Fahrenheit

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Incorrect Choices:

(A)
This wrong choice is $3110$ divided by $5$, which is just one step in the calculation. In order to get the right answer, you have to multiply this by $9$ and add to the result $32.$

(B)
Tip: Just because a number appears in the question doesn’t mean it is the answer.
This wrong choice is the boiling point of curium in degrees Celsius, as stated in the prompt. We're to convert it to degrees Fahrenheit.

(C)
If you forget to add $32$, like this

$F=\frac{9}{5}C\fbox{+32}=\frac{9}{5}\cdot 3110 = 9 \cdot 622 = 5598,$

you will get this wrong answer.

(E)
This answer choice is meant to confuse you.

The magnitude of the gravitational force acting on an object of mass $m$ located a distance $h$ above Earth's surface is

$$F_{g}=G\frac{M_{E}\cdot m}{(R_{E} +h)^{2}},$$

where $G=6.67 \times 10^{-11} \text{N}\cdot \text{m}^{2}/\text{kg}^{2}$ is the gravitational constant, $M_{E}=5.98 \times 10^{24}$ kg is the Earth's mass and $R_{E}=6.37 \times 10^6$ m is the Earth's radius. If the International Space Station has a mass of $4.31 \times 10^{5}$ kg and the station operates $3.5 \times 10^{5}$ m above the surface of the Earth, what is the gravitational force acting on it?

(A) $\ \ 8.83\ \text{N}$
(B) $\ \ 3.80\times 10^6\ \text{N}$
(C) $\ \ 4.24\ \times 10^6\ \text{N}$
(D) $\ \ 2.56\times 10^{13}\ \text{N}$
(E) $\ \ 5.71 \times 10^{16}\ \text{N}$

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Correct Answer: B

Solution:

We plug the given values into the equation:

$$\begin{array}{r c l} F_{g} &=& G\frac{M_{E}\cdot m}{(R_{E} +h)^{2}} \\ &=& 6.67 \times 10^{-11}\ \text{N}\cdot \text{m}^{2}/\text{kg}^{2} \times \frac{(5.98 \times 10^{24}\ \text{kg}) \cdot (4.31 \times 10^{5}\ \text{kg})}{(6.37 \times 10^6\ \text{m} + 3.5 \times 10^{5}\ \text{m})^{2}} \\ &=& 3.80\times 10^6\ \text{N} \\ \end{array}$$

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Incorrect Choices:

(A)
If you forget to plug $m$ into the formula, you will get this wrong answer.

(C)
If you forget to account for $h$, you will get this wrong answer.

(D)
If you forget to square the denominator, you will get this wrong answer.

(E)
If you forget $G$, you will get this wrong answer.

If you thought these examples difficult and you need to review the material, these links will help:

• SAT Translating Word Problems
• Exponential Functions-Growth and Decay
• Graphs of Exponential Functions
• Quadratic Equations Word Problems
• SAT Exponents
• Rules of Exponents

• Use a calculator.
• Exponential growth: $y=ba^{x}$, where $a>1$ and $b > 0.$
• Exponential decay: $y=ba^{x}$, where $0<a<1$ and $b>0.$
• Know the Rules of Exponents.
• Follow order of operations.
• SAT General Tips