Super-Hard SAT Advanced Math Question 21 | Free SAT Prep

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Question 21·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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For positive constants $a$ and $b$ , the function $f$ is defined by $f(x)=ab^x$ . It is known that

\begin{aligned} f(2)+f(3)&=72 \\ f(4)-f(3)&=108 \end{aligned}

The function $g$ is defined by $g(x)=f(2x-1)-6$ . Which choice gives an equation for $g$ ?

For exponential functions, values whose inputs differ by $1$ are related by multiplication by the same factor. When conditions involve several nearby inputs, name one output, such as $f(3)$ , and express the others using the common factor. Dividing the resulting equations often removes the unknown output and leaves an equation for the exponential factor. After finding the original function, carefully apply every part of an input transformation.

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Use a nearby output as a reference

Let $F=f(3)$ . Write $f(2)$ and $f(4)$ in terms of $F$ and the common exponential factor $b$ .

Eliminate one variable

After rewriting both conditions, divide one equation by the other. This removes $F$ and gives an equation involving only $b$ .

Track both parts of the input transformation

When substituting $2x-1$ into an exponential function, the $2x$ changes the base, while the $-1$ changes the coefficient.

Desmos Guide

Use algebra first

Algebra is faster because the two conditions can be divided to eliminate $f(3)$ . As a Desmos verification, graph the quadratic obtained after that division.

Verify the exponential factor

Enter $y=2x^2-5x-3$ . The positive $x$ -intercept is the allowable value of $b$ .

Verify the coefficient and transformation

Enter $54/3^3$ to confirm the coefficient $a$ . Then graph $y=2\cdot3^{2x-1}-6$ and $y=\frac{2}{3}\cdot9^x-6$ ; the graphs coincide.

Step-by-step Explanation

Express nearby function values using $f(3)$

Let $F=f(3)$ . Since each increase of $1$ in the input multiplies the output by $b$ ,

\begin{aligned} f(2)&=\frac{F}{b} \\ f(4)&=bF \end{aligned}

Therefore, the two given conditions become $F\left(1+\frac{1}{b}\right)=72$ and $F(b-1)=108$ .

Find the exponential factor

Divide the second equation by the first equation to eliminate $F$ :

\frac{b-1}{1+\frac{1}{b}}=\frac{108}{72}=\frac{3}{2}

This gives

2b^2-5b-3=0

Factoring yields $(2b+1)(b-3)=0$ . Because $b$ is positive, $b=3$ .

Find the original function

Using $F(b-1)=108$ with $b=3$ gives $F=54$ . Since $f(3)=a\cdot3^3$ ,

a\cdot27=54

so $a=2$ . Thus, $f(x)=2\cdot3^x$ .

Apply the transformation

Substitute $2x-1$ for the input of $f$ :

\begin{aligned} g(x)&=2\cdot3^{2x-1}-6 \\ &=\frac{2}{3}\cdot3^{2x}-6 \\ &=\frac{2}{3}\cdot9^x-6 \end{aligned}

This is choice D.

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