The function is defined by where is a positive constant, is a constant, and . If and , and for every positive integer , the value of is % less than the value of , which choice is the value of ? Which choice is the value of ?

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Super-Hard SAT Math Question 178 | Free SAT Prep | Aniko

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Question 178·200 Super-Hard SAT Math Questions·Advanced Math

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The function $f$ is defined by

f(x)=k\,b^{2x+1}+c

where $k$ is a positive constant, $c$ is a constant, and $0<b<1$ . If $f\left(-\tfrac12\right)=256$ and $f\left(\tfrac32\right)=16$ , and for every positive integer $n$ , the value of $f(n)$ is $p$ % less than the value of $f(n-1)$ , which choice is the value of $p$ ?

Which choice is the value of $p$ ?

When a problem says one term is a constant percent less than the previous term for all integers, immediately rewrite it as a constant ratio $f(n)=r f(n-1)$ . If the function includes an added constant (a vertical shift), use the “for all $n$ ” condition to test whether that shift must be $0$ . After that, use ratios of given function values to cancel unknown multipliers like $k$ , solve for the base, find the one-step ratio, and convert to percent decrease with $p=100(1-r)$ .

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Rewrite “ $p$ % less” as multiplication

If a quantity is $p$ % less than another, it equals $(1-\tfrac{p}{100})$ times the other. Call this constant multiplier $r$ .

Use the fact this happens for every positive integer $n$

Write $f(n)=r\,f(n-1)$ using $f(x)=k\,b^{2x+1}+c$ and simplify. If an expression must be true for all $n$ , terms that depend on $n$ can’t be cancelled by constants unless their coefficients are zero.

Once $c$ is determined, use a ratio of given values

After handling $c$ , divide $f\left(\tfrac32\right)$ by $f\left(-\tfrac12\right)$ to eliminate $k$ and get an equation in $b$ .

Desmos Guide

Model the constant ratio idea

In Desmos, define a constant ratio r (slider) to represent the fact that for integers $n$ , $f(n)=r f(n-1)$ .

Use the integer-step property to identify the ratio

Note that increasing $x$ by 1 increases the exponent $2x+1$ by 2, so the multiplicative factor from $f(n-1)$ to $f(n)$ is b^2 (this is the ratio r).

Solve for $b$ from the given values

Enter b^4 = 16/256 and restrict attention to $0<b<1$ to get the valid solution for b.

Convert the ratio to percent decrease

Enter 100*(1-b^2) to compute $p$ .

Step-by-step Explanation

Translate “ $p$ % less” into a constant ratio

If $f(n)$ is $p$ % less than $f(n-1)$ for every positive integer $n$ , then there is a constant multiplier $r$ such that

f(n)=r\,f(n-1)

for all positive integers $n$ , where $r=1-\frac{p}{100}$ and $0<r<1$ .

Use the constant-ratio condition to constrain $c$

Using the given formula for integer $n$ ,

k\,b^{2n+1}+c=r\bigl(k\,b^{2(n-1)+1}+c\bigr)=r\bigl(k\,b^{2n-1}+c\bigr).

Rearrange:

k\,b^{2n-1}(b^2-r)+c(1-r)=0.

As $n$ changes, $b^{2n-1}$ changes (since $0<b<1$ ), so for this to hold for all positive integers $n$ , both coefficients must be $0$ :

b^2-r=0 \quad\text{and}\quad c(1-r)=0.

Because $0<r<1$ , we have $1-r\ne 0$ , so $c=0$ and $r=b^2$ .

Use the two given values to solve for $b$

With $c=0$ , the function is $f(x)=k\,b^{2x+1}$ . Form the ratio:

\frac{f\left(\tfrac32\right)}{f\left(-\tfrac12\right)} =\frac{k\,b^{2\left(\tfrac32\right)+1}}{k\,b^{2\left(-\tfrac12\right)+1}} =b^{\left(2\left(\tfrac32\right)+1\right)-\left(2\left(-\tfrac12\right)+1\right)}.

Compute exponents:

2\left(\tfrac32\right)+1=4,\qquad 2\left(-\tfrac12\right)+1=0,

so the ratio is $b^4$ . Using the given function values,

b^4=\frac{16}{256}=\frac{1}{16}=\left(\frac12\right)^4,

and since $0<b<1$ , $b=\frac12$ .

Convert the ratio to a percent decrease

From Step 2, the one-step ratio is

\frac{f(n)}{f(n-1)}=r=b^2=\left(\frac12\right)^2=\frac14.

So $f(n)$ is $25\%$ of $f(n-1)$ , meaning it is $100\%-25\%=75\%$ less.

Therefore, $p$ is $75$ .

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Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation