Question 172·200 Super-Hard SAT Math Questions·Advanced Math

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For positive real numbers $m$ and $r$ with $m\ne 1$ , the following equations are true:

m^{\frac{4}{5}}=r^{-\frac{3}{10}}

m^{6c-2}=r^2

Which choice gives the value of $c$ ?

Two equations with powers of different bases call for the power-of-a-power rule. Raise one equation to make its power of $r$ match the other equation’s $r^2$ . Then equate the exponents on $m$ and use a Desmos regression to solve for $c$ . Keep track of the negative exponent: losing its sign changes the result.

Hints

Hint 1
The second equation has $r^2$ . In the first equation, $r$ has exponent $-\frac{3}{10}$ . A power of a power multiplies exponents, so what outside exponent would turn $-\frac{3}{10}$ into $2$ ?
Hint 2
Raise both sides of the first equation to that outside exponent. The exponent on $m$ changes too: multiply $\frac45$ by the outside exponent. What power of $m$ now equals $r^2$ ?
Hint 3
Both equations now give an expression for $r^2$ . Because $m$ is positive and not $1$ , equal powers of $m$ must have equal exponents. What equation does that give you for $c$ ?

Step-by-step

Match the powers of $r$
Step 1
Find the outside exponent
The second equation contains $r^2$ , so change $r^{-\frac{3}{10}}$ into that same power. In a power of a power, exponents multiply. Divide the target exponent by the current one, keeping its negative sign:
$2\div\left(-\frac{3}{10}\right)=-\frac{20}{3}.$
Step 2
Make $r^2$ appear
Raise both sides of the first equation to $-\frac{20}{3}$ :
$(m^{\frac45})^{-\frac{20}{3}}=(r^{-\frac{3}{10}})^{-\frac{20}{3}}.$
Multiply each pair of exponents:
$m^{(\frac45)(-\frac{20}{3})}=r^{(-\frac{3}{10})(-\frac{20}{3})}.$
Simplify the exponents:
$m^{-\frac{16}{3}}=r^2.$
Step 3
Equate the exponents on $m$
The second equation also equals $r^2$ , so set the two powers of $m$ equal:
$m^{6c-2}=m^{-\frac{16}{3}}.$
The given condition $m\ne1$ matters: if the positive base were $1$ , every exponent would give $1$ . Here, equal powers of $m$ must have equal exponents, so:
$6c-2=-\frac{16}{3}.$
Step 4
Solve for $c$ in Desmos
Type $6c-2\sim-\frac{16}{3}$ . The $\sim$ tells Desmos to find the value of $c$ that makes the sides equal. Under PARAMETERS, it reports $c\approx-0.555556$ ; the choices use exact fractions, so check that form next.
Step 5
Read the exact value
Type $c$ on the next line and tap the fraction button beside its value. Desmos displays $-\frac59$ , so $c=-\frac59$ . Choice A.

Question 172·200 Super-Hard SAT Math Questions·Advanced Math

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