A basic square numbers equation I couldn't solve this equation. Could you help me to solve this step by step.
Thank you.
$$\sqrt[4]{\frac{8^x}{8}}=\sqrt[3]{16}$$

\begin{align}
\sqrt[4]{8^{x} \times \frac{1}{8}} &= \sqrt[3]{16} \\ 
\sqrt[4]{8^{x-1}} &= \sqrt[3]{16} \\ 
8^{x-1} &= \left( \sqrt[3]{16} \right)^{4} \\ 
8^{x-1} &= 16^{\frac{4}{3}} \\ 
2^{3x-3} &= 2^{\frac{16}{3}} \\ 
3x-3 &= \frac{16}{3} \quad \to \quad
3x = \frac{16}{3} +3 \quad \to \quad
x = 1+\frac{16}{9} \quad \to \quad
x = \frac{25}{9} 
\end{align}
