False.Solution:
$LHS \Big(\frac{2}{3}\Big)^{-2}\times\Big(\frac{2}{3}\Big)^{-5}$
Using law of exponents, $a^m \times a^n=(a)^{m+n}[\because$ a is non-zero integer$]$
$\therefore$ $\Big(\frac{2}{3}\Big)^{-2}\times\Big(\frac{2}{3}\Big)^{-5}=\Big(\frac{2}{3}\Big)^{-2-5}$
$=\Big(\frac{2}{3}\Big)^{-7}$
$LHS ≠ RHS$
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