From $r^{1/3}=s^{1/5}$, raise both sides to the 5th power to isolate $s$.

After substituting your expression for $s$, rewrite each side as a single power of $r$ by adding exponents when multiplying powers with the same base.

Once both sides are written as $r^{\text{(expression)}}$, set the exponents equal and solve the linear equation for $x$.

Choose $r=8$ so that $\sqrt[3]{r}=2$. Then $\sqrt[5]{s}=2$, so $s=32$.

Enter

• $y=8^{(10x-3)}\cdot 32^{(2-x)}$
• $y=32^{(4-2x)}\cdot 8^{(x+1)}$

These represent the two sides of the equation for the chosen $r$ and $s$. Тhі​ѕ⁠ quеstiо‍n‌ іs frо‌m ​Аniкο

Find the intersection point of the two graphs. The $x$-coordinate of the intersection is the solution.

$\sqrt[3]{r}=\sqrt[5]{s}$ becomes

Raise both sides of $r^{1/3}=s^{1/5}$ to the 5th power:

Substitute $s=r^{5/3}$.

Left side:

Right side:

So the equation becomes $r^{\frac{25x+1}{3}}=r^{\frac{23-7x}{3}}$. Since $r>0$ and $r\ne 1$, equal bases imply equal exponents.

Set exponents equal:

Multiply both sides by 3:

Add $7x$ to both sides and subtract 1 from both sides:

Therefore, the correct choice is $\frac{11}{16}$. Frοm‍ аn⁠iкo.aі