Let For this exponential function , let . Which choice rewrites so that appears as the coefficient (the number multiplying the exponential term)?

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Super-Hard SAT Math Question 17 | Free SAT Prep | Aniko

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Question 17·200 Super-Hard SAT Math Questions·Advanced Math

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Let

f(x)=\frac{4^{x+1}\cdot 9^{2x-3}}{6^{3x-5}}.

For this exponential function $f$ , let $k=f(-1)$ . Which choice rewrites $f(x)$ so that $k$ appears as the coefficient (the number multiplying the exponential term)? ©‌ anік⁠ο.аi

When a function is built from several exponentials with different bases, rewrite each base in primes (often $2$ and $3$ ), then combine exponents to get a clean form like $C\left(\frac{3}{2}\right)^x$ . To make $f(a)$ appear as a coefficient, rewrite the function as $f(a)\cdot b^{x-a}$ so plugging in $x=a$ makes the exponent $0$ and leaves the coefficient. Роwerе‍d bу Аniко

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Use the same prime bases

Rewrite $4$ , $9$ , and $6$ using only powers of $2$ and $3$ so you can combine exponents.

Combine powers separately

After rewriting, collect all factors of $2$ together and all factors of $3$ together, then subtract exponents when dividing. А⁠nik‌о.аі - ‍SАT ‍Pr‍e‌p

Make the input value create an exponent of 0

To show $f(-1)$ as a coefficient, look for a form where the exponent becomes $0$ when $x=-1$ .

Desmos Guide

Enter the original function

Type

f(x)=(4^(x+1)*9^(2x-3))/(6^(3x-5))

as a function.

Evaluate $k=f(-1)$

In a new line, type k=f(-1) and note the value Desmos displays for k.

Check which option shows that value as a coefficient

For each answer choice, look at its coefficient (the number multiplying the exponential). The correct choice is the one whose coefficient exactly matches the value you found for k. © Anікο

Step-by-step Explanation

Rewrite using prime factors

Rewrite each base using $2$ and $3$ :

$4=2^2$
$9=3^2$
$6=2\cdot 3$

\begin{aligned} 4^{x+1} &= 2^{2x+2}\\ 9^{2x-3} &= 3^{4x-6}\\ 6^{3x-5} &= 2^{3x-5}\cdot 3^{3x-5} \end{aligned}

Combine exponents

Substitute and combine powers of 2 and 3:

\begin{aligned} f(x) &= \frac{2^{2x+2}\cdot 3^{4x-6}}{2^{3x-5}\cdot 3^{3x-5}}\\ &=2^{(2x+2)-(3x-5)}\cdot 3^{(4x-6)-(3x-5)}\\ &=2^{-x+7}\cdot 3^{x-1} \end{aligned}

Factor out constants and rewrite as a single exponential ratio:

\begin{aligned} 2^{-x+7}\cdot 3^{x-1} &=2^7\cdot 2^{-x}\cdot 3^{-1}\cdot 3^x\\ &=\frac{128}{3}\left(\frac{3}{2}\right)^x \end{aligned}

Find $k=f(-1)$

Now evaluate using the simplified form:

\begin{aligned} k=f(-1) &= \frac{128}{3}\left(\frac{3}{2}\right)^{-1}\\ &=\frac{128}{3}\cdot\frac{2}{3}\\ &=\frac{256}{9} \end{aligned}

Choose the form that displays $k$ as the coefficient

A form with exponent $(x+1)$ will have its coefficient equal to $f(-1)$ , because substituting $x=-1$ makes the exponent $0$ .

The only choice whose coefficient is $\frac{256}{9}=k$ is:

$f(x)=\frac{256}{9}\left(\frac{3}{2}\right)^{x+1}$ A⁠nі⁠кo AІ Т‍utοr

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