The equation below relates distinct positive real numbers , , and . Which choice correctly expresses in terms of and ?

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Super-Hard SAT Math Question 106 | Free SAT Prep | Aniko

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Question 106·200 Super-Hard SAT Math Questions·Advanced Math

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The equation below relates distinct positive real numbers $p$ , $q$ , and $r$ . Which choice correctly expresses $r$ in terms of $p$ and $q$ ?

p = 6q \cdot \sqrt[3]{\left(\frac{r}{25}\right)^{-4}}

When solving for a variable inside a radical or power, rewrite every radical as a rational exponent first, then isolate the powered expression. If a negative exponent appears, flip the fraction to make the exponent positive. Finally, “undo” the remaining exponent by raising both sides to its reciprocal, and only at the end multiply/divide to clear any constant factors like $25$ .

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Convert the cube root to an exponent

Use $\sqrt[3]{x}=x^{1/3}$ and combine the exponents when you have something like $(\cdot)^{-4}$ inside the radical.

Deal with the negative exponent

A negative exponent means you can flip the fraction inside: $(A)^{-k}=\left(\frac{1}{A}\right)^k$ .

Undo a power of $\frac{4}{3}$

If $X=Y^{4/3}$ , what power should you raise both sides to in order to get $Y$ by itself?

Desmos Guide

Create sliders for constants

Create sliders for $p$ and $q$ (choose positive values, for example $p=10$ and $q=2$ ).

Model the original equation as a function of $r$

Define

$L=p$
$R(r)=6q\left(\left(\frac{r}{25}\right)^{-4}\right)^{1/3}$

Then graph $y=L$ and $y=R(r)$ (use $x$ as the input variable if needed: graph $y=R(x)$ ).

Compute candidate values of $r$ from the answer choices

Define four expressions for $r$ using the choices, such as:

$r_1=25\left(\frac{p}{6q}\right)^{3/4}$
$r_2=25\left(\frac{6q}{p}\right)^{3/4}$
$r_3=25\left(\frac{6q}{p}\right)^{4/3}$
$r_4=\frac{25}{6q}p^{3/4}$

(Enter each exactly as written.)

Check which candidate makes the equation true

For each candidate $r_i$ , compute the difference

D_i = p-6q\left(\left(\frac{r_i}{25}\right)^{-4}\right)^{1/3}.

The correct choice will give a value of $D_i$ that is $0$ (or extremely close due to rounding).

Identify the matching option

The candidate that makes $D_i=0$ corresponds to the correct expression for $r$ , which matches $25\left(\frac{6q}{p}\right)^{\frac{3}{4}}$ .

Step-by-step Explanation

Rewrite the radical as an exponent

Since $\sqrt[3]{x}=x^{1/3}$ ,

\sqrt[3]{\left(\frac{r}{25}\right)^{-4}}=\left(\frac{r}{25}\right)^{-4\cdot \frac{1}{3}}=\left(\frac{r}{25}\right)^{-\frac{4}{3}}.

So the equation becomes

\frac{p}{6q}=\left(\frac{r}{25}\right)^{-\frac{4}{3}}.

Remove the negative exponent by taking reciprocals

A negative exponent means reciprocal:

\left(\frac{r}{25}\right)^{-\frac{4}{3}}=\left(\frac{25}{r}\right)^{\frac{4}{3}}.

Taking reciprocals of both sides gives

\frac{6q}{p}=\left(\frac{r}{25}\right)^{\frac{4}{3}}.

Undo the exponent $\frac{4}{3}$

Raise both sides to the reciprocal power $\frac{3}{4}$ :

\left(\frac{6q}{p}\right)^{\frac{3}{4}}=\frac{r}{25}.

Multiply both sides by $25$ :

r=25\left(\frac{6q}{p}\right)^{\frac{3}{4}}.

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Step-by-step Explanation

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Step-by-step Explanation