Question 70·200 Super-Hard SAT Math Questions·Advanced Math

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Two numbers $r$ and $s$ are each greater than zero. Suppose

\sqrt[4]{r}=\sqrt[7]{s}

and

r^2s^3=2^{29}.

Which choice is equal to $\sqrt[14]{r^5s^4}$ ?

Equal roots of two positive numbers are a cue to write both numbers as powers of one shared base. Use the product condition to make an equation for that base, then solve it with a Desmos regression. Finally, use a rational exponent, a fraction used as a power, to rewrite the requested root. Keep the root's index separate from the exponent in the product condition.

Hints

Hint 1
The equal fourth and seventh roots have one shared positive value. Call it $x$ . Raising $x$ to the fourth power gives $r$ ; raising it to the seventh gives $s$ .
Hint 2
For a power of a power, multiply the exponents: $(x^4)^2=x^8$ . Then, because the bases match, multiplying $x^8$ and $x^{21}$ means adding their exponents. What equation does the given product become?
Hint 3
A fourteenth root gives the whole product an exponent of $1/14$ . First write $r^5s^4$ as one power of the shared base. Which exponent must you divide by $14$ ?

Step-by-step

Use the shared root as a base
Step 1
Write both numbers with one base
Let $x=\sqrt[4]{r}=\sqrt[7]{s}$ . A root undoes a power, so $r=x^4$ and $s=x^7$ . The given positivity means $x>0$ . Equal roots give both numbers the same base.
Step 2
Put the shared base into the product
Replace $r$ with $x^4$ and $s$ with $x^7$ in the given product: $r^2s^3=2^{29}\quad\Longrightarrow\quad (x^4)^2(x^7)^3=2^{29}$ .
Step 3
Raise each power
For a power of a power, multiply the exponents, so $4\cdot2=8$ and $7\cdot3=21$ : $(x^4)^2(x^7)^3=x^8x^{21}$ .
Step 4
Combine the matching bases
Multiplying powers of the same base adds their exponents: $x^8x^{21}=x^{8+21}=x^{29}$ . So the product condition is $x^{29}=2^{29}$ .
Step 5
Find the positive shared base
Type $x_1^{29}\sim2^{29}\{x_1>0\}$ . Desmos uses $\sim$ to find $x_1$ ; the restriction keeps the positive value the problem requires. Under PARAMETERS, it reports $x_1=2$ , so $x=2$ .
Step 6
Rewrite the requested root
A fourteenth root raises the entire product to $1/14$ . Replace $r$ and $s$ with their powers of $x$ : $\sqrt[14]{r^5s^4}=\big((x^4)^5(x^7)^4\big)^{1/14}$ .
Step 7
Raise the powers inside
Multiply each pair of exponents inside the root: $\big((x^4)^5(x^7)^4\big)^{1/14}=(x^{20}x^{28})^{1/14}$ .
Step 8
Combine the powers inside
Both factors have base $x$ , so add their exponents: $(x^{20}x^{28})^{1/14}=(x^{20+28})^{1/14}=(x^{48})^{1/14}$ .
Step 9
Apply the fourteenth root
The root divides the exponent by $14$ , not by the $29$ from the other equation: $(x^{48})^{1/14}=x^{48/14}=2^{48/14}$ . Type $48/14$ , then $x_1^{48/14}$ . Desmos shows $3.428571\ldots$ for the exponent; its fraction button shows $24/7$ . It also shows $10.7672015\ldots$ for $x_1^{48/14}$ , the requested root's approximate value. Since $x_1=2$ , the requested root is $2^{24/7}$ . Choice A.

Question 70·200 Super-Hard SAT Math Questions·Advanced Math

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