A function can be written in the form , where , , and are real constants with and . Suppose and . What is the value of ?

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SAT Nonlinear Functions Question 33 | Free SAT Prep | Aniko

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Question 33·Hard·Nonlinear Functions

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A function can be written in the form $f(x) = a b^x + c$ , where $a$ , $b$ , and $c$ are real constants with $a \ne 0$ and $b > 1$ . Suppose $f(1) - f(0) = 12$ and $f(3) - f(2) = 108$ . What is the value of $b$ ?

For exponential functions of the form $f(x)=ab^x+c$ , differences over equal intervals often factor nicely. Write each given difference symbolically (like $f(1)-f(0)$ and $f(3)-f(2)$ ), simplify and factor to expose common factors, then use the given numerical values to form equations. Taking the ratio of these equations is a powerful shortcut: it cancels out unwanted constants like $a$ and $c$ and leaves a simple equation in $b$ alone, which you can then solve quickly and match to the choices.

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Write the function values explicitly

Start by writing $f(1)$ , $f(0)$ , $f(3)$ , and $f(2)$ using the formula $f(x)=ab^x+c$ . Then subtract to find $f(1)-f(0)$ and $f(3)-f(2)$ in terms of $a$ , $b$ , and $c$ .

Notice what cancels in the differences

When you compute $f(1)-f(0)$ and $f(3)-f(2)$ , pay attention to which terms cancel out. Can you factor each difference to see a common factor?

Use ratios to eliminate unknowns

Once you have expressions for both differences that each include a common factor, set them equal to the given numbers (12 and 108). How can dividing one equation by the other help you solve for $b$ without knowing $a$ or $c$ ?

Remember the condition on $b$

After you get an equation involving $b^2$ , you will have two possible values for $b$ . Use the fact that $b>1$ to decide which one is valid.

Desmos Guide

Use Desmos to compute the key ratio

After you derive (by hand) that $b^2 = \dfrac{108}{12}$ , type 108/12 into Desmos. Note the value Desmos gives you; that is the value of $b^2$ .

Use Desmos to find $b$ from $b^2$

In Desmos, type sqrt(108/12) to get the positive square root, which is the value of $b$ that satisfies $b>1$ . Compare this value to the answer choices to select the matching option.

Step-by-step Explanation

Express the given differences using the function formula

The function is $f(x)=ab^x+c$ .

$f(1)=ab^1+c=ab+c$
$f(0)=ab^0+c=a+c$

So the first difference is

f(1)-f(0)=(ab+c)-(a+c)=a(b-1).

Similarly,

$f(3)=ab^3+c$
$f(2)=ab^2+c$

So the second difference is

f(3)-f(2)=(ab^3+c)-(ab^2+c)=ab^3-ab^2=ab^2(b-1).

Use the given numerical values to form equations

We are told:

$f(1)-f(0)=12$
$f(3)-f(2)=108$

From Step 1:

$a(b-1)=12$
$ab^2(b-1)=108$

So we have the system

\begin{aligned} a(b-1) &= 12, \\ ab^2(b-1) &= 108. \end{aligned}

Eliminate $a$ and $(b-1)$ by dividing the equations

Divide the second equation by the first:

\frac{ab^2(b-1)}{a(b-1)} = \frac{108}{12}.

The $a$ and $(b-1)$ cancel, leaving

b^2 = \frac{108}{12}.

Now you only need to evaluate this fraction and solve for $b$ .

Solve for $b$ and use the condition $b>1$

Compute the fraction:

\frac{108}{12}=9,

b^2 = 9.

The solutions to this equation are $b=3$ and $b=-3$ , but the problem states $b>1$ , so we must take $b=3$ .

Thus, the correct answer choice is A) 3.

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Step-by-step Explanation

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Step-by-step Explanation