Question

If n³ = x and n² = 20x, where n > 0, what is the value of x? ​

Answer 1

$n^2 = 20x\implies |n| = \sqrt{20x}$

As $n > 0$ the previous equation is equivalent to $n=\sqrt{20x}$ with the condition that $x > 0$.

Plugging that into $n^3 = x$ we have

$20^{3/2}\cdot x^{3/2} = x\iff x\cdot\Big(20^{3/2}\sqrt{x}-1\Big) = 0$, where $x > 0$.

Therefore,

$20^{3/2}\cdot\sqrt{x} - 1 = 0 \iff \sqrt{x} = 20^{-3/2} \iff x = 20^{-3}$.

Ans. $x=20^{-3} = 0.000125$