| x | f(x) | |---|---| | 1 | | | 2 | | | 4 | | For the exponential function , the table above shows selected values of and the corresponding values of , where is a constant greater than . Assume that can be written in the form for some constants , , and . If is a positive number such that , what is the value of ?

Solution available with step-by-step explanation

SAT Nonlinear Functions Question 51 | Free SAT Prep | Aniko

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Question 51·Hard·Nonlinear Functions

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x	f(x)
1	$b^{2}$
2	$b^{7}$
4	$b^{23}$

For the exponential function $f$ , the table above shows selected values of $x$ and the corresponding values of $f(x)$ , where $b$ is a constant greater than $1$ .

Assume that $f$ can be written in the form $f(x)=b^{ax^{2}+px+c}$ for some constants $a$ , $p$ , and $c$ . If $k$ is a positive number such that $f(k)=b^{14}$ , what is the value of $k$ ?

Your answer

(Express the answer as an integer)

For problems where an exponential function is written with a common base, immediately translate given function values into equations by equating exponents. Use elimination or substitution to solve for the unknown parameters in the exponent (here, $a$ , $p$ , and $c$ ). Once you have the explicit formula for the exponent, turn any condition like $f(k)=b^{\text{(number)}}$ into a simple equation by setting the exponent equal to that number, then solve the resulting linear or quadratic equation, keeping only solutions that satisfy any given conditions (such as being positive).

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Use the fact that the base is the same

Each value of $f(x)$ is a power of the same base $b$ . When $b>1$ , if $b^m = b^n$ then $m=n$ . Use this idea to write equations for the exponents at $x=1,2$ , and $4$ .

Create and simplify a system for $a$ , $p$ , and $c$

From the three $x$ -values in the table, write three equations in $a$ , $p$ , and $c$ . Then use subtraction (elimination) to reduce the system and solve for $a$ and $p$ .

Find $f(x)$ , then set up an equation for $k$

Once you know $a$ , $p$ , and $c$ , write the exponent as a quadratic in $x$ . Then use $f(k)=b^{14}$ to set up an equation by equating the exponent to $14$ , and solve that quadratic for the positive value of $k$ .

Desmos Guide

Graph the quadratic equation for $k$

In Desmos, enter the equation y = k^2 + 2k - 15, but replace k with x (Desmos uses $x$ as the variable), so type y = x^2 + 2x - 15. This graphs the quadratic whose roots are the possible values of $k$ .

Find the positive solution for $k$

Use Desmos to find the $x$ -intercepts of the graph (tap or click where the curve crosses the $x$ -axis). You will see one negative and one positive $x$ -intercept; the positive $x$ -intercept is the value of $k$ that satisfies $f(k)=b^{14}$ .

Step-by-step Explanation

Translate the table into equations for the exponent

We are told that $f(x)=b^{ax^2+px+c}$ and given three values of $f(x)$ .

From the table:

When $x=1$ , $f(1)=b^2$ , so

a(1)^2 + p(1) + c = 2

which simplifies to $a + p + c = 2$ .

When $x=2$ , $f(2)=b^7$ , so

4a + 2p + c = 7

When $x=4$ , $f(4)=b^{23}$ , so

16a + 4p + c = 23

We now have a system of three equations in $a$ , $p$ , and $c$ :
(1) $a + p + c = 2$
(2) $4a + 2p + c = 7$
(3) $16a + 4p + c = 23$ .

Solve the system to find $a$ and $p$

Subtract equation (1) from equation (2):

\begin{aligned} (4a + 2p + c) - (a + p + c) &= 7 - 2 \\ 3a + p &= 5 \end{aligned}

Subtract equation (2) from equation (3):

\begin{aligned} (16a + 4p + c) - (4a + 2p + c) &= 23 - 7 \\ 12a + 2p &= 16 \end{aligned}

Divide the last equation by 2:

6a + p = 8

Now subtract $(3a + p = 5)$ from $(6a + p = 8)$ :

\begin{aligned} (6a + p) - (3a + p) &= 8 - 5 \\ 3a &= 3 \end{aligned}

So $a = 1$ . Substitute $a=1$ into $3a + p = 5$ :

\begin{aligned} 3(1) + p &= 5 \\ 3 + p &= 5 \\ p &= 2 \end{aligned}

So we have $a=1$ and $p=2$ .

Find $c$ and write the formula for $f(x)$

Use equation (1), $a + p + c = 2$ , with $a=1$ and $p=2$ :

\begin{aligned} 1 + 2 + c &= 2 \\ 3 + c &= 2 \\ c &= -1 \end{aligned}

So the exponent is

ax^2 + px + c = 1\cdot x^2 + 2x - 1 = x^2 + 2x - 1.

Therefore the function is

f(x) = b^{x^2 + 2x - 1}.

Set up the equation for $k$ using $f(k)=b^{14}$

We are told that $f(k)=b^{14}$ . Using the formula we found,

f(k) = b^{k^2 + 2k - 1}.

Because the base $b$ is the same and $b>1$ , the exponents must be equal when the powers are equal:

k^2 + 2k - 1 = 14.

Move 14 to the left to get a standard quadratic equation:

\begin{aligned} k^2 + 2k - 1 - 14 &= 0 \\ k^2 + 2k - 15 &= 0 \end{aligned}

Now we just need to solve this quadratic for $k$ .

Solve the quadratic and choose the positive solution

Solve

k^2 + 2k - 15 = 0.

Factor the quadratic:

k^2 + 2k - 15 = (k + 5)(k - 3).

Set each factor equal to zero:

$k + 5 = 0 \Rightarrow k = -5$
$k - 3 = 0 \Rightarrow k = 3$

The problem states that $k$ is a positive number, so we discard $k=-5$ and keep $k=3$ .

Thus, the value of $k$ is 3.

Strategy

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Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation