Answer
$\frac{81}{16t^{4}}$
Work Step by Step
According to the definition of negative exponents, $a^{-n}=\frac{1}{a^{n}}$ (where $a\ne0$).
Therefore, $(\frac{2t}{3})^{-4}=\frac{1}{(\frac{2t}{3})^{4}}=(\frac{3}{2t})^{4}=\frac{3^{4}}{2^{4}t^{4}}=\frac{81}{16t^{4}}$.
