Tolong bantuan darurat, please

[tex]
\left(\frac {1}{8} \right)^{\frac{4}{3}}=\sqrt {16^{3x-5}}\\

(8^{-1})^{\frac{4}{3}}=\sqrt {2^{{(4)}^{(3x-5)}}}\\

(2^{-3})^{\frac{4}{3}}=2^{{(2)}^{(3x-5)}}\\

-4 = 6x - 10\\
6x = 10 - 4\\
6x = 6\\
x = 1
[/tex]