A basic square numbers equation

Question

I couldn't solve this equation. Could you help me to solve this step by step. Thank you.

$$\sqrt[4]{\frac{8^x}{8}}=\sqrt[3]{16}$$

Answer 1

\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}

Answer 2

$$2^{(3x-3)/4}=2^{4/3}$$ $$9x-9=16$$ $$x=25/9$$