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

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

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

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 $y=R(x)$).

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.)

For each candidate $r_i$, compute the difference

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

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

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

So the equation becomes

A negative exponent means reciprocal: Anіkο Qu​еstiοn ⁠Bаn‌k

Taking reciprocals of both sides gives

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

Multiply both sides by $25$: