If \(\sqrt{\sqrt{\sqrt{5x}}}=\sqrt[6]{4x},\) what is the range of the possible values of \(x?\)

What is the range of the possible value of x?
$$\left(5x\right)^{\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}}=\left(4x\right)^{\frac{1}{6}}$$
$$\left(5x\right)^{\frac{1}{8}}=\left(4x\right)^{\frac{1}{6}}$$
$$5x=\left[\left(4x\right)^{\frac{1}{6}}\right]^8$$
$$5x=\left(4x\right)^{\frac{8}{6}}$$
$$5x=\left(4x\right)^{\frac{4}{3}}$$
$$5x=4^{\frac{4}{3}}\cdot x^{\frac{4}{3}}$$
$$\frac{5x}{x\cdot4^{\frac{4}{3}}}=\frac{\left(4^{\frac{4}{3}}\cdot x^{\frac{4}{3}}\right)}{x\cdot4^{\frac{4}{3}}}$$
$$\frac{5}{4^{\frac{4}{3}}}=\frac{x^{\frac{4}{3}}}{x^1}$$ $$\frac{5}{4^{\frac{4}{3}}}=x^{\frac{4}{3}-1}$$ $$\frac{5}{4^{\frac{4}{3}}}=x^{\frac{4-3}{3}}$$
$$x^{\frac{1}{3}}=\frac{5}{4^{\frac{4}{3}}}$$
$$\left(x^{\frac{1}{3}}\right)^3=\left(\frac{5}{4^{\frac{4}{3}}}\right)^3$$
$$x=\frac{5^3}{4^{\frac{4}{3}\cdot3}}=\frac{5^3}{4^4}=\frac{125}{256}$$

Answer = C